Algebra - System of Equations and Inequalities

Solve the following system of equations: $\;$ $u + 2v = 2, \;\; \left|2u - 3v\right| = 1$

Given system of equations: $\;$ $u + 2v = 2$ $\;\;\; \cdots \; (1)$

$\left|2u - 3v\right| = 1$ $\;\;\; \cdots \; (2)$

When $\;$ $2u - 3v > 0$, $\;$ i.e. $\;$ when $\;$ $2u > 3v$, $\;$ $\left|2u - 3v\right| = 2u - 3v$

Then equation $(2)$ becomes

$2u - 3v = 1$ $\;\;\; \cdots \; (3)$

Solving equations $(1)$ and $(3)$ simultaneously gives

$-7v = -3$ $\implies$ $v = \dfrac{3}{7}$

Substituting $v = \dfrac{3}{7}$ in equation $(1)$ gives

$u + \dfrac{6}{7} = 2$ $\;$ i.e. $\;$ $u = 2 - \dfrac{6}{7} = \dfrac{8}{7}$

Now, when $u = \dfrac{8}{7}$ and $v = \dfrac{3}{7}$,

$2u = \dfrac{16}{7}$, $3v = \dfrac{9}{7}$ and the condition $2u > 3v$ is satisfied.

When $\;$ $2u - 3v < 0$, $\;$ i.e. $\;$ when $\;$ $2u < 3v$, $\;$ $\left|2u - 3v\right| = -2u + 3v$

Then equation $(2)$ becomes

$-2u + 3v = 1$ $\;\;\; \cdots \; (4)$

Solving equations $(1)$ and $(4)$ simultaneously gives

$7v = 5$ $\implies$ $v = \dfrac{5}{7}$

Substituting $v = \dfrac{5}{7}$ in equation $(1)$ gives

$u + \dfrac{10}{7} = 2$ $\;$ i.e. $\;$ $u = 2 - \dfrac{10}{7} = \dfrac{4}{7}$

Now, when $u = \dfrac{4}{7}$ and $v = \dfrac{5}{7}$,

$2u = \dfrac{8}{7}$, $3v = \dfrac{15}{7}$ and the condition $2u < 3v$ is satisfied.

$\therefore \;$ The solution to the given pair of equations is $\;$ $\left(u, v\right) = \left\{\left(\dfrac{8}{7}, \dfrac{3}{7}\right), \; \left(\dfrac{4}{7}, \dfrac{5}{7}\right) \right\}$

Algebra - System of Equations and Inequalities

Solve the following system of equations: $\;$ $x + 2y - 6 = 0, \;\; \left|x - 3\right| - y = 0$

Given system of equations: $\;$ $x + 2y - 6 = 0$ $\;\;\; \cdots \; (1)$

$\left|x - 3\right| - y = 0$ $\;\;\; \cdots \; (2)$

When $\;$ $x - 3 > 0$, $\;$ i.e. $\;$ when $\;$ $x > 3$, $\;$ $\left|x - 3\right| = x - 3$

Then equation $(2)$ becomes

$x - 3 - y = 0$ $\;\;\; \cdots \; (3)$

Solving equations $(1)$ and $(3)$ simultaneously gives

$3x - 12 = 0$ $\implies$ $x = 4$

The condition for this solution is $\;$ $x > 3$

and $\;$ $x = 4 > 3$

$\implies$ $x = 4$ $\;$ is a valid solution.

Substituting the value of $x$ in equation $(1)$ gives

$4 + 2y - 6 = 0$ $\;$ i.e. $\;$ $2y = 2$ $\;$ i.e. $\;$ $y = 1$

When $\;$ $x - 3 < 0$, $\;$ i.e. $\;$ when $\;$ $x < 3$, $\;$ $\left|x - 1\right| = - \left(x - 3\right) = 3 - x$

Then $(2)$ becomes

$3 - x - y = 0$ $\;\;\; \cdots \; (4)$

Solving equations $(1)$ and $(4)$ simultaneously gives

$x = 0$

The condition for this solution is $\;$ $x < 3$

and $\;$ $x = 0 < 3$

$\implies$ $x = 0$ $\;$ is a valid solution.

Substituting the value of $x$ in equation $(1)$ gives

$2y - 6 = 0$ $\implies$ $y = 3$

$\therefore \;$ The solution to the given pair of equations is $\;$ $\left(x, y\right) = \left\{\left(4, 1\right), \; \left(0, 3\right) \right\}$

Algebra - System of Equations and Inequalities

Solve the following system of equations: $\;$ $\left|x - 1\right| + y = 0, \;\; 2x - y = 1$

Given system of equations: $\;$ $\left|x - 1\right| + y = 0$ $\;\;\; \cdots \; (1)$

$2x - y = 1$ $\;\;\; \cdots \; (2)$

When $\;$ $x - 1 > 0$, $\;$ i.e. $\;$ when $\;$ $x > 1$, $\;$ $\left|x - 1\right| = x - 1$

Then $(1)$ becomes

$x - 1 + y = 0$ $\;\;$ i.e. $\;$ $x + y = 1$ $\;\;\; \cdots \; (3)$

Solving equations $(2)$ and $(3)$ simultaneously gives

$3x = 2$ $\implies$ $x = \dfrac{2}{3}$

The condition for this solution is $\;$ $x > 1$

But $\;$ $x = \dfrac{2}{3} < 1$

$\implies$ $x = \dfrac{2}{3}$ $\;$ is not a valid solution.

When $\;$ $x - 1 < 0$, $\;$ i.e. $\;$ when $\;$ $x < 1$, $\;$ $\left|x - 1\right| = - \left(x - 1\right) = 1 - x$

Then $(1)$ becomes

$1 - x + y = 0$ $\;$ i.e. $\;$ $x - y = 1$ $\;\;\; \cdots \; (4)$

Solving equations $(2)$ and $(4)$ simultaneously gives

$x = 0$

The condition for this solution is $\;$ $x < 1$

and $\;$ $x = 0 < 1$

$\implies$ $x = 0$ $\;$ is a valid solution.

Substituting the value of $x$ in equation $(2)$ gives

$-y = 1$ $\implies$ $y = -1$

$\therefore \;$ The solution to the given pair of equations is $\;$ $\left(x, y\right) = \left\{\left(0, -1\right) \right\}$

Algebra - System of Equations and Inequalities

Solve the following system of equations: $\;$ $x + 3 \left|y\right| - 1 = 0, \;\; x + y + 3 = 0$

Given system of equations: $\;$ $x + 3 \left|y\right| - 1 = 0$ $\;\;\; \cdots \; (1)$

$x + y + 3 = 0$ $\;\;\; \cdots \; (2)$

When $\;$ $y > 0$, $\;$ $\left|y\right| = + y$

Then $(1)$ becomes

$x + 3y - 1 = 0$ $\;\;\; \cdots \; (3)$

Solving equations $(2)$ and $(3)$ simultaneously gives

$2y - 4 = 0$ $\implies$ $y = 2$

Substituting $\;$ $y = 2$ $\;$ in equation $(2)$ gives

$x + 2 + 3 = 0$ $\implies$ $x = -5$

When $\;$ $y < 0$, $\;$ $\left|y\right| = -y$

Then $(1)$ becomes

$x - 3y - 1 = 0$ $\;\;\; \cdots \; (4)$

Solving equations $(2)$ and $(4)$ simultaneously gives

$-4y - 4 = 0$ $\implies$ $y = -1$

Substituting the value of $y$ in equation $(2)$ gives

$x - 1 + 3 = 0$ $\implies$ $x = -2$

$\therefore \;$ The required solution is $\;$ $\left(x, y\right) = \left\{\left(-5, 2\right), \; \left(-2, -1\right) \right\}$

Algebra - System of Equations and Inequalities

Solve the following system of equations: $\;$ $y + x - 1 = 0, \; \left|y\right| - x - 1 = 0$

Given system of equations: $\;$ $y + x - 1 = 0$ $\;\;\; \cdots \; (1)$

$\left|y\right| - x - 1 = 0$ $\;\;\; \cdots \; (2)$

When $\;$ $y > 0$, $\;$ $\left|y\right| = + y$

Then $(2)$ becomes

$y - x - 1 = 0$ $\;\;\; \cdots \; (3)$

Adding equations $(1)$ and $(3)$ gives

$2y - 2 = 0$ $\implies$ $y = 1$

Substituting $\;$ $y = 1$ $\;$ in equation $(1)$ gives

$x = 1 - 1 = 0$

When $\;$ $y < 0$, $\;$ $\left|y\right| = -y$

Then $(2)$ becomes

$-y - x - 1 = 0$

i.e. $\;$ $x + y = -1$ $\;\;\; \cdots \; (4)$

From equation $(1)$, $\;$ $x + y = 1$ $\;\;\; \cdots \; (1a)$

$\therefore \;$ When $\;$ $\left|y\right| = - y$, $\;$ we see from equations $(1a)$ and $(4)$ that equations $(1)$ and $(4)$ cannot be solved simultaneously.

$\therefore \;$ The required solution is $\;$ $\left(x, y\right) = \left\{\left(0, 1\right) \right\}$

Algebra - Equations and Inequations

Solve the equation: $\;$ $\left(x - 1\right) \left(x - 2\right) \left(x - 3\right) \left(x - 4\right) = 15$

Given equation: $\;$ $\left(x - 1\right) \left(x - 2\right) \left(x - 3\right) \left(x - 4\right) = 15$

i.e. $\;$ $\left(x^2 - 5x + 4\right) \left(x^2 - 5x + 6\right) = 15$ $\;\;\; \cdots \; (1)$

Let $\;$ $x^2 - 5x = p$ $\;\;\; \cdots \; (2)$

Then $(1)$ becomes

$\left(p +4\right) \left(p + 6\right) = 15$

i.e. $\;$ $p^2 + 10p + 24 = 15$

i.e. $\;$ $p^2 + 10p + 9 = 0$

i.e. $\;$ $\left(p + 9\right) \left(p + 1\right) = 0$

i.e. $\;$ $p = -9$ $\;$ or $\;$ $p = -1$

When $\;$ $p = -9$, $(2)$ becomes

$x^2 - 5x = -9$

i.e. $\;$ $x^2 - 5x + 9 = 0$ $\;\;\; \cdots \; (3)$

Discriminant of $(3)$ is

$\Delta = \left(-5\right)^2 - 4 \times 1 \times 9 = = 25 - 36 = - 9 < 0$

$\implies$ $(3)$ does not have any real roots.

When $\;$ $p = -1$, $\;$ $(2)$ becomes

$x^2 - 5x = -1$

i.e. $\;$ $x^2 - 5x + 1 = 0$ $\;\;\; \cdots \; (4)$

Solving $(4)$ for $x$ gives

$x = \dfrac{5 + \sqrt{21}}{2}$ $\;$ or $\;$ $x = \dfrac{5 - \sqrt{21}}{2}$

$\therefore \;$ Solution of the given equation is

$x = \left\{\dfrac{5 + \sqrt{21}}{2}, \; \dfrac{5 - \sqrt{21}}{2} \right\}$

Algebra - Equations and Inequations

Solve the equation: $\;$ $\left(x - \sqrt{3}\right)^4 - 5 \left(x - \sqrt{3}\right)^2 + 4 = 0$

Given equation: $\;$ $\left(x - \sqrt{3}\right)^4 - 5 \left(x - \sqrt{3}\right)^2 + 4 = 0$ $\;\;\; \cdots \; (1)$

Let $\;$ $\left(x - \sqrt{3}\right)^2 = p$ $\;\;\; \cdots \; (2)$

Then $(1)$ becomes

$p^2 - 5p + 4 = 0$

i.e. $\;$ $\left(p - 4\right) \left(p - 1\right) = 0$

i.e. $\;$ $p = 4$ $\;$ or $\;$ $p = 1$

When $\;$ $p = 4$, $\;$ $(2)$ becomes

$\left(x - \sqrt{3}\right)^2 = 4$

i.e. $\;$ $x - \sqrt{3} = \pm 2$

i.e. $\;$ $x = \sqrt{3} + 2$ $\;$ or $\;$ $x = \sqrt{3} - 2$

When $\;$ $p = 1$, $\;$ $(2)$ becomes

$\left(x - \sqrt{3}\right)^2 = 1$

i.e. $\;$ $x - \sqrt{3} = \pm 1$

i.e. $\;$ $x = \sqrt{3} + 1$ $\;$ or $\;$ $x = \sqrt{3} - 1$

$\therefore \;$ The solutions of $(1)$ are

$x = \left\{\sqrt{3} + 2, \sqrt{3} - 2, \sqrt{3} + 1, \sqrt{3} - 1 \right\}$

Algebra - Equations and Inequations

Solve the inequation: $\;$ $x^2 - \left|5x - 3\right| - x < 2$

Given inequation: $\;$ $x^2 - \left|5x - 3\right| - x < 2$ $\;\;\; \cdots \; (1)$

Case 1:

When $\;$ $5x - 3 > 0$

i.e. $\;$ when $\;$ $x > \dfrac{3}{5}$, $\;\;$ $\left|5x - 3\right| = 5x - 3$ Then $(1)$ becomes

$x^2 - \left(5x - 3\right) -x < 2$

i.e. $\;$ $x^2 - 6x + 1 < 0$ $\;\;\; \cdots \; (2)$

Solving the quadratic equation $\;$ $x^2 - 6x + 1 = 0$ $\;$ gives

$x = 3 + 2 \sqrt{2}$ $\;\;\; \cdots \; (3a)$ $\;$ OR $\;$ $x = 3 - 2 \sqrt{2}$ $\;\;\; \cdots \; (3b)$

$\therefore \;$ In view of $(3a)$ and $(3b)$, $(2)$ becomes

$\left[x - \left(3 + 2 \sqrt{2}\right)\right] \left[x - \left(3 - 2 \sqrt{2}\right)\right] < 0$

i.e. $\;$ $x - 3 - 2 \sqrt{2} < 0$ $\;$ AND $\;$ $x - 3 + 2 \sqrt{2} > 0$

OR $\;$ $x - 3 - 2 \sqrt{2} > 0$ $\;$ AND $\;$ $x - 3 + 2 \sqrt{2} < 0$

i.e. $\;$ $x < 3 + 2 \sqrt{2}$ $\;$ AND $\;$ $x > 3 - 2 \sqrt{2}$ $\implies$ $x \in \left(3 - 2 \sqrt{2}, 3 + 2 \sqrt{2}\right)$

OR $\;$ $x > 3 + 2 \sqrt{2}$ $\;$ AND $\;$ $x < 3 - 2 \sqrt{2}$ $\;$ which is not possible

$\therefore \;$ When $\;$ $5x - 3 > 0$, i.e. when $x > \dfrac{3}{5}$, $\;$ solution of $(1)$ is $\;$ $x \in \left(3 - 2 \sqrt{2}, 3 + 2 \sqrt{2}\right)$ $\;\;\; \cdots \; (4)$

Case 2:

When $\;$ $5x - 3 < 0$

i.e. $\;$ when $\;$ $x < \dfrac{3}{5}$, $\;$ $\left|5x - 3\right| = -5x + 3$

Then $(1)$ becomes

$x^2 - \left(-5x + 3\right) - x < 2$

i.e. $\;$ $x^2 + 4x - 5 < 0$

i.e. $\;$ $\left(x + 5\right) \left(x - 1\right) < 0$

i.e. $\;$ $x + 5 < 0$ $\;$ AND $\;$ $x - 1 > 0$

OR $\;$ $x + 5 > 0$ $\;$ AND $\;$ $x - 1 < 0$

i.e. $\;$ $x < -5$ $\;$ AND $\;$ $x > 1$ $\;\;$ which is not possible

OR $\;$ $x > -5$ $\;$ AND $\;$ $x < 1$ $\implies$ $x \in \left(-5, 1\right)$

$\therefore \;$ When $\;$ $5x - 3 < 0$, i.e. when $x < \dfrac{3}{5}$, $\;$ solution of $(1)$ is $\;$ $x \in \left(-5, 1\right)$ $\;\;\; \cdots \; (5)$

$\therefore \;$ In view of $(4)$ and $(5)$, solution of $(1)$ is $\;$ $x \in \left(-5, 3 + 2 \sqrt{2}\right)$

Algebra - Equations and Inequations

Solve the inequation: $\;$ $x^2 - 7x + 12 < \left|x - 4\right|$

Given inequation: $\;$ $x^2 - 7x + 12 < \left|x - 4\right|$ $\;\;\; \cdots \; (1)$

Case 1:

When $\;$ $x - 4 > 0$

i.e. $\;$ when $\;$ $x > 4$, $\;\;$ $\left|x - 4\right| = x - 4$

Then $(1)$ becomes

$x^2 - 7x + 12 < x - 4$

i.e. $\;$ $x^2 - 8x + 16 < 0$

i.e. $\;$ $\left(x - 4\right)^2 < 0$

i.e. $\;$ $x - 4 < 0$ $\implies$ $x < 4$

$\therefore \;$ When $\;$ $x - 4 > 0$, i.e. when $x > 4$ $\;$ solution of $(1)$ is $\;$ $x < 4$ $\;\;\; \cdots \; (2)$

Case 2:

When $\;$ $x - 4 < 0$

i.e. $\;$ when $\;$ $x < 4$, $\;$ $\left|x - 4\right| = -x + 4$

Then $(1)$ becomes

$x^2 - 7x + 12 < -x + 4$

i.e. $\;$ $x^2 - 6x + 8 < 0$

i.e. $\;$ $\left(x - 2\right) \left(x - 4\right) < 0$

i.e. $\;$ $x - 2 < 0$ $\;$ AND $\;$ $x - 4 > 0$

OR $\;$ $x - 2 > 0$ $\;$ AND $\;$ $x - 4 < 0$

i.e. $\;$ $x < 2$ $\;$ AND $\;$ $x > 4$ $\;\;$ which is not possible

OR $\;$ $x > 2$ $\;$ AND $\;$ $x < 4$ $\implies$ $x \in \left(2, 4\right)$

$\therefore \;$ When $\;$ $x - 4 < 0$, i.e. when $x < 4$, $\;$ solution of $(1)$ is $\;$ $x \in \left(2, 4\right)$ $\;\;\; \cdots \; (3)$

$\therefore \;$ In view of $(2)$ and $(3)$, solution of $(1)$ is $\;$ $x \in \left(2, 4\right)$

Algebra - Equations and Inequations

Solve the inequation: $\;$ $\left|x^2 - 2x\right| < x$

Given inequation: $\;$ $\left|x^2 - 2x\right| < x$ $\;\;\; \cdots \; (1)$

Case 1:

When $\;$ $x^2 - 2x > 0$

i.e. $\;$ when $\;$ $x \left(x - 2\right) > 0$

i.e. $\;$ when $\;$ $x > 0 \; \cap \; x - 2 > 0$ $\;\;$ OR $\;\;$ $x < 0 \; \cap \; x - 2 < 0$

i.e. $\;$ when $\;$ $x > 0 \; \cap \; x > 2$ $\;\;$ OR $\;\;$ $x < 0 \; \cap \; x < 2$

i.e. $\;$ when $\;$ $x > 2$ $\;$ OR $\;$ $x < 2$, $\;\;$ $\left|x^2 - 2x\right| = x^2 - 2x$

Then $(1)$ becomes

$x^2 - 2x < x$

i.e. $\;$ $x^2 - 3x < 0$

i.e. $\;$ $x \left(x - 3\right) < 0$

i.e. $\;$ $x < 0 \; \cap \; x - 3 > 0$ $\;$ OR $\;$ $x > 0 \; \cap \; x - 3 < 0$

i.e. $\;$ $x < 0 \; \cap \; x > 3$ $\;\;$ which is not possible

OR $\;$ $x > 0 \; \cap \; x < 3$ $\implies$ $x \in \left(0, 3\right)$

$\therefore \;$ When $\;$ $x^2 - 2x > 0$, $\;$ $x \in \left(0, 3\right)$ $\;\;\; \cdots \; (2)$

Case 2:

When $\;$ $x^2 - 2x < 0$

i.e. $\;$ when $\;$ $x \left(x - 2\right) < 0$

i.e. $\;$ when $\;$ $x < 0 \; \cap \; x - 2 > 0$ $\;\;$ OR $\;\;$ $x > 0 \; \cap \; x - 2 < 0$

i.e. $\;$ when $\;$ $x < 0 \; \cap \; x > 2$ $\;\;$ which is not possible

OR $\;\;$ $x > 0 \; \cap \; x < 2$ $\implies$ $x \in \left(0,2\right)$

i.e. $\;$ when $\;$ $x \in \left(0, 2\right)$, $\;\;$ $\left|x^2 - 2x\right| = -x^2 + 2x$

Then $(1)$ becomes

$-x^2 + 2x < x$

i.e. $\;$ $-x^2 + x < 0$

i.e. $\;$ $x^2 - x > 0$

i.e. $\;$ $x \left(x - 1\right) > 0$

i.e. $\;$ $x < 0 \; \cap \; x - 1 < 0$ $\;$ OR $\;$ $x > 0 \; \cap \; x - 1 > 0$

i.e. $\;$ $x < 0 \; \cap \; x < 1$ $\;$ OR $\;$ $x > 0 \; \cap \; x > 1$

i.e. $\;$ $x < 0$ $\;$ OR $\;$ $x > 1$ $\implies$ $x \in \left(x < 0 \; \cup \; x > 1\right)$

$\therefore \;$ When $\;$ $x^2 - 2x < 0$, $\;$ $x \in \left(x < 0 \; \cup \; x > 1\right)$ $\;\;\; \cdots \; (3)$

$\therefore \;$ In view of $(2)$ and $(3)$, solution of $(1)$ is $\;\;$ $x \in \left(1, 3\right)$

Algebra - Equations and Inequations

Solve the inequation: $\;$ $\left|x^2 + x\right| - 5 < 0$

Given inequation: $\;$ $\left|x^2 + x\right| - 5 < 0$ $\;\;\; \cdots \; (1)$

Case 1:

When $\;$ $x^2 + x > 0$

i.e. $\;$ when $\;$ $x \left(x + 1\right) > 0$

i.e. $\;$ when $\;$ $x > 0 \; \cap \; x + 1 > 0$ $\;\;$ OR $\;\;$ $x < 0 \; \cap \; x + 1 < 0$

i.e. $\;$ when $\;$ $x > 0 \; \cap \; x > -1$ $\;\;$ OR $\;\;$ $x < 0 \; \cap \; x < -1$

$\therefore \;$ when $\;$ $x > 0$ $\;$ OR $\;$ $x < -1$, $\;\;$ $\left|x^2 + x\right| = x^2 + x$

Then $(1)$ becomes

$x^2 + x - 5 < 0$ $\;\;\; \cdots \; (2)$

Solving the quadratic equation $\;\;$ $x^2 + x - 5 = 0$ $\;\;\; \cdots \; (3)$ $\;$ gives

$x = \dfrac{-1 \pm \sqrt{1^2 - 4 \times 1 \times \left(-5\right)}}{2 \times 1}$

i.e. $\;$ $x = \dfrac{-1+ \sqrt{21}}{2}$ $\;\;\; \cdots \; (4a)$ $\;$ OR $\;$ $x = \dfrac{-1 - \sqrt{21}}{2}$ $\;\;\; \cdots \; (4b)$

In view of $(4a)$ and $(4b)$, $(2)$ can be written as

i.e. $\;$ $\left(x - \dfrac{1 - \sqrt{21}}{2}\right) \left(x - \dfrac{1 + \sqrt{21}}{2}\right) < 0$

i.e. $\;$ $x - \dfrac{1 - \sqrt{21}}{2} < 0 \; \cap \; x - \dfrac{1 + \sqrt{21}}{2} > 0$

OR $\;\;$ $x - \dfrac{1 - \sqrt{21}}{2} > 0 \; \cap \; x - \dfrac{1 + \sqrt{21}}{2} < 0$

i.e. $\;$ $x < \dfrac{1 - \sqrt{21}}{2} \; \cap \; x > \dfrac{1 + \sqrt{21}}{2}$ $\;\;$ which is not possible

OR $\;\;$ $x > \dfrac{1 - \sqrt{21}}{2} \; \cap \; x < \dfrac{1 + \sqrt{21}}{2}$ $\implies$ $x \in \left(\dfrac{1 - \sqrt{21}}{2}, \dfrac{1 + \sqrt{21}}{2}\right)$

$\therefore \;$ When $\;$ $x^2 + x > 0$, $\;$ solution of $(1)$ is $\;\;\;$ $x \in \left(\dfrac{1 - \sqrt{21}}{2}, \dfrac{1 + \sqrt{21}}{2}\right)$ $\;\;\; \cdots \; (5)$

Case 2:

When $\;$ $x^2 + x < 0$

i.e. $\;$ when $\;$ $x \left(x + 1\right) < 0$

i.e. $\;$ when $\;$ $x < 0 \; \cap \; x + 1 > 0$ $\;\;$ OR $\;\;$ $x > 0 \; \cap \; x + 1 < 0$

i.e. $\;$ when $\;$ $x < 0 \; \cap \; x > -1$ $\implies$ $x \in \left(-1, 0\right)$

OR $\;\;$ $x > 0 \; \cap \; x < -1$ $\;\;$ which is not possible

$\therefore \;$ when $\;$ $x \in \left(-1, 0\right)$, $\;\;$ $\left|x^2 + x\right| = -x^2 - x$

then $(1)$ becomes

$-x^2 - x - 5 < 0$

i.e. $\;$ $x^2 + x + 5 > 0$ $\;\;\; \cdots \; (6)$

Consider the quadratic equation $\;\;$ $x^2 + x + 5 = 0$ $\;\;\; \cdots \; (7)$

Discriminant of $(7)$ is $\;\;$ $\Delta = 1^2 - 4 \times 1 \times 5 = - 19 < 0$

$\implies$ Equation $(7)$ and hence inequality $(6)$ have no real solution.

$\therefore \;$ Solution of $(1)$ is

$x \in \left(\dfrac{1 - \sqrt{21}}{2}, \dfrac{1 + \sqrt{21}}{2}\right)$ $\;$ when $\;$ $x > 0$ $\;$ OR $\;$ $x < -1$

Algebra - Equations and Inequations

Solve the inequation: $\;$ $\left|x^2 - 4x\right| < 5$

Given inequation: $\;$ $\left|x^2 - 4x\right| < 5$ $\;\;\; \cdots \; (1)$

Case 1:

When $\;$ $x^2 - 4x > 0$

i.e. $\;$ when $\;$ $x \left(x - 4\right) > 0$

i.e. $\;$ when $\;$ $x > 0 \; \cap \; x - 4 > 0$ $\;\;$ OR $\;\;$ $x < 0 \; \cap \; x - 4 < 0$

i.e. $\;$ when $\;$ $x > 0 \; \cap \; x > 4$ $\implies$ $x > 4$

OR $\;\;$ $x < 0 \; \cap \; x < 4$ $\implies$ $x < 0$

$\therefore \;$ When $\;$ $x < 0$ $\;$ OR $\;$ $x > 4$, $\;\;$ $\left|x^2 - 4x\right| = x^2 - 4x$

Then $(1)$ becomes

$x^2 - 4x < 5$

i.e. $\;$ $x^2 - 4x - 5 < 0$

i.e. $\;$ $\left(x - 5\right) \left(x + 1\right) < 0$

i.e. $\;$ $x - 5 < 0 \; \cap \; x + 1 > 0$ $\;\;$ OR $\;\;$ $x - 5 > 0 \; \cap \; x + 1 < 0$

i.e. $\;$ $x < 5 \; \cap \; x > -1$ $\implies$ $x \in \left(-1, 5\right)$

OR $\;\;$ $x > 5 \; \cap \; x < -1$ $\;\;$ which is not possible

$\therefore \;$ When $\;$ $x < 0$ $\;$ OR $\;$ $x > 4$, $\;$ solution of $(1)$ is $\;\;\;$ $x \in \left(-1, 5\right)$ $\;\;\; \cdots \; (2)$

Case 2:

When $\;$ $x^2 - 4x < 0$

i.e. $\;$ when $\;$ $x \left(x - 4\right) < 0$

i.e. $\;$ when $\;$ $x < 0 \; \cap \; x - 4 > 0$ $\;\;$ OR $\;\;$ $x > 0 \; \cap \; x - 4 < 0$

i.e. $\;$ when $\;$ $x < 0 \; \cap \; x > 4$ $\;\;$ which is not possible

OR $\;\;$ $x > 0 \; \cap \; x < 4$ $\implies$ $x \in \left(0, 4\right)$

$\therefore \;$ When $\;$ $x \in \left(0, 4\right)$, $\;\;$ $\left|x^2 - 4x\right| = -x^2 + 4x$

Then $(1)$ becomes

$-x^2 + 4x < 5$

i.e. $\;$ $x^2 - 4x + 5 < 0$

The equation $\;$ $x^2 - 4x + 5 = 0$ $\;$ does not have any real roots since its discriminant $\Delta$ is

$\Delta = \left(-4\right)^2 - 4 \times 1 \times 5 = 16 - 20 = -4 < 0$

$\therefore \;$ From $(2)$, solution of $(1)$ is $\;$ $x \in \left(-1, 5\right)$ $\;$ when $\;$ $x < 0$ $\;$ OR $\;$ $x > 4$

Algebra - Equations and Inequations

Solve the inequation: $\;$ $x^2 - 5 \left|x\right| + 6 < 0$

Given inequation: $\;$ $x^2 - 5 \left|x\right| + 6 < 0$ $\;\;\; \cdots \; (1)$

When $\;$ $x > 0$, $\;$ $\left|x\right| = + x$

Then $(1)$ becomes

$x^2 - 5x + 6 < 0$

i.e. $\;$ $\left(x - 3\right) \left(x - 2\right) < 0$

i.e. $\;$ $x - 3 < 0 \; \cap \; x - 2 > 0$ $\;\;$ OR $\;\;$ $x - 3 > 0 \; \cap \; x - 2 < 0$

i.e. $\;$ $x < 3 \; \cap \; x > 2$ $\implies$ $x \in \left(2, 3\right)$

OR $\;\;$ $x > 3 \; \cap \; x < 2$ $\;\;$ which is not possible

$\therefore \;$ When $x > 0$, solution of $(1)$ is $\;\;\;$ $x \in \left(2,3\right)$ $\;\;\; \cdots \; (2)$

When $\;$ $x < 0$, $\;$ $\left|x\right| = - x$

Then $(1)$ becomes

$x^2 + 5x + 6 < 0$

i.e. $\;$ $\left(x + 2\right) \left(x + 3\right) < 0$

i.e. $\;$ $x + 2 < 0 \; \cap \; x + 3 > 0$ $\;\;$ OR $\;\;$ $x + 2 > 0 \; \cap \; x + 3 < 0$

i.e. $\;$ $x < -2 \; \cap \; x > -3$ $\implies$ $x \in \left(-3, -2\right)$

OR $\;\;$ $x > -2 \; \cap \; x < -3$ $\;\;$ which is not possible

$\therefore \;$ When $x < 0$, solution of $(1)$ is $\;\;\;$ $x \in \left(-3, -2\right)$ $\;\;\; \cdots \; (3)$

$\therefore \;$ From $(2)$ and $(3)$, solution of $(1)$ is $\;$ $x \in \left(-3, -2\right) \; \cup \; \left(2, 3\right)$

Algebra - Equations and Inequations

Solve the inequation: $\;$ $2 - x - x^2 \geq 0$

Given inequation: $\;$ $2 - x - x^2 \geq 0$

i.e. $\;$ $x^2 + x - 2 \leq 0$

i.e. $\;$ $\left(x - 1\right) \left(x + 2\right) \leq 0$

i.e. $\;$ $x - 1 \leq 0 \; \cap \; x + 2 \geq 0$ $\;$ OR $\;$ $x - 1 \geq 0 \; \cap \; x + 2 \leq 0$

i.e. $\;$ $x \leq 1 \; \cap \; x \geq -2$ $\implies$ $x \in \left[-2, 1\right]$

OR $\;$ $x \geq 1 \; \cap \; x \leq -2$ $\;\;\;$ which is not possible

$\therefore \;$ The solution is $\;$ $x \in \left[-2, 1\right]$

Algebra - Equations and Inequations

Solve the inequation: $\;$ $x^2 - 14x - 15 > 0$

Given inequation: $\;$ $x^2 - 14x - 15 > 0$

i.e. $\;$ $\left(x - 15\right) \left(x + 1\right) > 0$

i.e. $\;$ $x - 15 > 0 \; \cap \; x + 1 > 0$ $\;$ OR $\;$ $x - 15 < 0 \; \cap \; x + 1 < 0$

i.e. $\;$ $x > 15 \; \cap \; x > -1$ $\implies$ $x > 15$

OR $\;$ $x < 15 \; \cap \; x < -1$ $\implies$ $x < -1$

i.e. $\;$ $x \in \left(15, + \infty\right)$ $\;$ OR $\;$ $x \in \left(- \infty, -1\right)$

$\therefore \;$ The solution is $\;$ $x \in \left(- \infty, -1\right) \cup \left(15, + \infty\right)$

Algebra - Equations and Inequations

Solve the inequation: $\;$ $3x^2 - 7x -6 < 0$

Given inequation: $\;$ $3x^2 - 7x - 6 < 0$

i.e. $\;$ $\left(3x + 2\right) \left(x - 3\right) < 0$

i.e. $\;$ $3x + 2 < 0 \; \cap \; x - 3 > 0$ $\;$ OR $\;$ $3x + 2 > 0 \; \cap \; x - 3 < 0$

i.e. $\;$ $x < \dfrac{-2}{3} \; \cap \; x > 3$ $\;$ which is not possible

OR $\;$ $x > \dfrac{-2}{3} \; \cap \; x < 3$ $\implies$ $x \in \left(\dfrac{-2}{3}, 3\right)$

$\therefore \;$ The solution is $\;$ $x \in \left(\dfrac{-2}{3}, 3\right)$

Algebra - Equations and Inequations

Solve the inequation: $\;$ $3x^2 - 7x + 4 \leq 0$

Given inequation: $\;$ $3x^2 - 7x + 4 \leq 0$

i.e. $\;$ $\left(3x - 4\right) \left(x - 1\right) \leq 0$

i.e. $\;$ $3x - 4 \leq 0 \; \cap \; x - 1 \geq 0$ $\;$ OR $\;$ $3x - 4 \geq 0 \; \cap \; x - 1 \leq 0$

i.e. $\;$ $x \leq \dfrac{4}{3} \; \cap \; x \geq 1$ $\implies$ $x \in \left[1, \dfrac{4}{3}\right]$

OR $\;$ $x \geq \dfrac{4}{3} \; \cap \; x \leq 1$ $\;$ which is not possible

$\therefore \;$ The solution is $\;$ $x \in \left[1, \dfrac{4}{3}\right]$

Algebra - Equations and Inequations

Calculate $\dfrac{1}{x_1^3} + \dfrac{1}{x_2^3}$, where $x_1$ and $x_2$ are the roots of the equation $2x^2 - 3ax -2 = 0$.

Given quadratic equation: $\;$ $2x^2 - 3ax - 2 = 0$

The roots of the given quadratic equation are $\;$ $x_1$ $\;$ and $\;$ $x_2$.

Sum of roots $= x_1 + x_2 = \dfrac{3a}{2}$ $\;\;\; \cdots \; (1a)$

Product of roots $= x_1 \cdot x_2 = \dfrac{-2}{2} = -1$ $\;\;\; \cdots \; (1b)$

Now,

$\begin{aligned} \dfrac{1}{x_1^3} + \dfrac{1}{x_2^3} & = \dfrac{x_1^3 + x_2^3}{x_1^3 \cdot x_2^3} \\\\ & = \dfrac{\left(x_1 + x_2\right) \left[x_1^2 - x_1 \cdot x_2 + x_2^2\right]}{x_1^3 \cdot x_2^3} \\\\ & = \dfrac{\left(x_1 + x_2\right) \left[\left(x_1 + x_ 2\right)^2 - 3 x_1 \cdot x_2\right]}{\left(x_1 \cdot x_2\right)^3} \\\\ & = \dfrac{\dfrac{3a}{2} \left[\left(\dfrac{3a}{2}\right)^2 - 3 \times \left(-1\right)\right]}{\left(-1\right)^3} \;\;\; \left[\text{by equations (1a) and (1b)}\right] \\\\ & = \dfrac{-3a}{2} \left[\dfrac{9a^2}{4} + 3\right] \\\\ & = \dfrac{-27a^3}{8} - \dfrac{9a}{2} \end{aligned}$

Algebra - Equations and Inequations

Express $x_1^3 + x_2^3$ in terms of the coefficients of the equation $x^2 + px + q = 0$, where $x_1$ and $x_2$ are the roots of the equation.

Given quadratic equation: $\;$ $x^2 + px + q = 0$

Roots of the given quadratic equation are $\;$ $x_1$ $\;$ and $x_2$.

Sum of roots $= x_1 + x_2 = -p$ $\;\;\; \cdots \; (1a)$

Product of roots $= x_1 \cdot x_2 = q$ $\;\;\; \cdots \; (1b)$

Now,

$\begin{aligned} x_1^3 + x_2^3 & = \left(x_1 + x_2\right) \left[x_1^2 - x_1 \cdot x_2 + x_2^2\right] \\\\ & = \left(x_1 + x_2\right) \left[\left(x_1 + x_2\right)^2 - 2x_1 \cdot x_2 - x_1 \cdot x_2\right] \\\\ & = \left(x_1 + x_2\right) \left[\left(x_1 + x_2\right)^2 - 3 x_1 \cdot x_2\right] \;\;\; \cdots \; (2) \end{aligned}$

In view of equations $(1a)$ and $(1b)$, equation $(2)$ becomes

$\begin{aligned} x_1^3 + x_2^3 & = \left(-p\right) \left[\left(-p\right)^2 - 3q\right] \\\\ & = 3pq - p^3 \end{aligned}$

Algebra - Equations and Inequations

Given two quadratic equations $x^2 - x + m = 0$ and $x^2 - x + 3m = 0$, $m \neq 0$. Find the value of $m$ for which one of the roots of the second equation is equal to double the root of the first equation.

Given quadratic equations: $\;$ $x^2 - x + m = 0$ $\;\;\; \cdots \; (1)$

$x^2 - x + 3m = 0$ $\;\;\; \cdots \; (2)$

Let $\alpha_1$ and $\beta_1$ be the roots of equation $(1)$.

Let $\alpha_2$ and $\beta_2$ be the roots of equation $(2)$.

As per question, $\;$ $\alpha_2 = 2 \alpha_1$ $\;\;\; \cdots \; (3)$

Now, $\;$ $\alpha_1 + \beta_1 = -1$ $\;\;\; \cdots \; (4a)$; $\;\;$ $\alpha_1 \cdot \beta_1 = m$ $\;\;\; \cdots \; (4b)$

$\alpha_2 + \beta_2 = -1$ $\;\;$ i.e. $\;\;$ $2 \alpha_1 + \beta_2 = -14$ $\;\;\; \cdots \; (5a)$ $\;\;\;$ [in view of (3)]

and $\;$ $\alpha_2 \cdot \beta_2 = 3m$ $\;\;$ i.e. $\;\;$ $2 \alpha_1 \cdot \beta_2 = 3m$ $\;\;\; \cdots \; (5b)$ $\;\;\;$ [in view of (3)]

We have from equations $(4a)$ and $(5a)$,

$2 \beta_1 - \beta_2 = -1$ $\;\;\; \cdots \; (6a)$

From equation $(4b)$, $\;$ $\alpha_1 = \dfrac{m}{\beta_1}$

From equation $(5b)$, $\;$ $\alpha_1 = \dfrac{3m}{2 \beta_2}$

$\therefore \;$ We have, $\;$ $\dfrac{m}{\beta_1} = \dfrac{3m}{2 \beta_2}$

i.e. $\;$ $\beta_2 = \dfrac{3}{2} \beta_1$ $\;\;\; \cdots \; (6b)$ $\;\;$ provided $\;$ $m \neq 0$

In view of equation $(6b)$, equation $(6a)$ becomes

$2 \beta_1 - \dfrac{3}{2} \beta_1 = -1$

i.e. $\;$ $\beta_1 = -2$

Substituting the value of $\beta_1$ in equation $(4b)$ gives

$\alpha_1 = -1 - \beta_1 = -1 + 2 = 1$

Substituting the values of $\alpha_1$ and $\beta_1$ in equation $(4b)$ gives

$1 \times \left(-2\right) = m$ $\implies$ $m = -2$

Algebra - Equations and Inequations

For what values of $\;$ $a$ $\;$ do the equations $x^2 + ax + 1 = 0$ and $x^2 + x + a = 0$ have a root in common?

Given quadratic equations: $\;$ $x^2 + ax + 1 = 0$ $\;\;\; \cdots \; (1)$

$x^2 + x + a = 0$ $\;\;\; \cdots \; (2)$

Let $\alpha$ be the common root of equations $(1)$ and $(2)$.

Then $\alpha$ satisfies both equations $(1)$ and $(2)$.

$\therefore \;$ We have,

$\alpha^2 + a \alpha + 1 = 0$ $\;\;\; \cdots \; (3)$; $\;\;$ $\alpha^2 + \alpha + a = 0$ $\;\;\; \cdots \; (4)$

$\implies$ $\alpha^2 + a \alpha + 1 = \alpha^2 + \alpha + a$

i.e. $\;$ $a \alpha + 1 = \alpha + a$ $\;\;$ provided $\;$ $\alpha \neq 0$

i.e. $\;$ $\alpha \left(a - 1\right) = a - 1$

i.e. $\;$ $\alpha = 1$ $\;$ if $\;$ $a - 1 = 0$ $\;$ i.e. $\;$ $a \neq 1$

Substituting the value of $\alpha$ in equation $(3)$ gives

$1 + a + 1 = 0$

i.e. $\;$ $a = -2$

Algebra - Equations and Inequations

Find the coefficients of the equation $x^2 + px + q = 0$ such that its roots are equal to $p$ and $q$.

Given quadratic equation: $\;$ $x^2 + px + q = 0$

Roots of the given quadratic equation are $\;$ $p, \; q$.

Sum of roots $= p + q = -p$ $\;\;\; \cdots \; (1)$

Product of roots $= p \cdot q = q$ $\;\;\; \cdots \; (2)$

From equation $(2)$, when $q \neq 0$, $\;$ $p = 1$

Substituting $\;$ $p = 1$ $\;$ in equation $(1)$ gives

$1 + q = -1$ $\implies$ $q = -2$

When $\;$ $q = 0$, $\;$ we have from $(1)$

$p + 0 = -p$

i.e. $\;$ $2p = 0$ $\implies$ $p = 0$

$\therefore \;$ Values of $p$ and $q$ are

$p = 0, \; q = 0$ $\;\;$ OR $\;\;$ $p = 1, \; q = -2$

Algebra - Equations and Inequations

For what values of $\;$ $a$ $\;$ do the graphs of the functions $\;$ $y = 2ax + 1$ $\;$ and $\;$ $y = \left(a - 6\right) x^2 - 2$ $\;$ do not intersect?

The given functions are: $\;$ $y = 2ax + 1$ $\;\;\; \cdots \; (1a)$ $\;$ and

$y = \left(a - 6\right) x^2 - 2$ $\;\;\; \cdots \; (1b)$

When functions $(1a)$ and $(1b)$ intersect, we have,

$2ax + 1 = \left(a - 6\right) x^2 - 2$

i.e. $\;$ $\left(a - 6\right) x^2 - 2ax - 3 = 0$ $\;\;\; \cdots \; (2)$

Now, $(1a)$ and $(1b)$ will not intersect when equation $(2)$ does not have any real roots

i.e. $\;$ when discriminant $\Delta$ of $(2)$ is less than $0$.

i.e. $\;$ Discriminant $\Delta = \left(-2a\right)^2 - 4 \times \left(a - 6\right) \times \left(-3\right) < 0$

i.e. $\;$ $4a^2 + 12a - 72< 0$

i.e. $\;$ $a^2 + 3a - 18 < 0$

i.e. $\;$ $\left(a - 3\right) \left(a + 6\right) < 0$

i.e. $\;$ $a < 3$ $\;$ and $\;$ $a > -6$ $\;\;\;$ OR $\;\;\;$ $a > 3$ $\;$ and $\;$ $a < -6$

i.e. $\;$ $-6 < a < 3$ $\;\;\;$ OR $\;\;\;$ Condition $a > 3 \cap a < -6 = \phi$ (null set)

$\therefore \;$ Curves $(1a)$ and $(1b)$ will not intersect when $a \in \left(-6, 3\right)$.

Algebra - Equations and Inequations

Find $\;$ $p$ $\;$ in the equation $\;$ $x^2 - 4x + p = 0$ $\;$ if it is known that the sum of the squares of its roots is equal to $16$.

Given quadratic equation: $\;\;\;$ $x^2 - 4x + p = 0$

Relation between the roots $\alpha$ and $\beta$ of the given quadratic equation is

$\alpha^2 + \beta^2 = 16$ $\;\;\; \cdots \; (1)$

Sum of roots of the given quadratic equation: $\;$ $\alpha + \beta = 4$ $\;\;\; \cdots \; (2a)$

Product of roots of the given quadratic equation: $\;$ $\alpha \cdot \beta = p$ $\;\;\; \cdots \; (2b)$

Squaring equation $(2a)$ gives

$\left(\alpha + \beta\right)^2 = 4^2$

i.e. $\;$ $\alpha^2 + \beta^2 + 2 \alpha \beta = 16$ $\;\;\; \cdots \; (3)$

In view of equations $(1)$ and $(2b)$, equation $(3)$ becomes

$16 + 2p = 16$

i.e. $\;$ $2p = 0$ $\implies$ $p = 0$

Algebra - Equations and Inequations

Find $\;$ $b$ $\;$ in the equation $\;$ $5x^2 + bx - 28 = 0$ $\;$ if the roots $x_1$ and $x_2$ of the equation are related as $5x_1 + 2x_2 = 1$ and $b$ is an integer.

Given quadratic equation: $\;\;\;$ $5x^2 + bx - 28 = 0$

Relation between the roots $x_1$ and $x_2$ of the given quadratic equation is

$5x_1 + 2x_2 = 1$ $\;\;\; \cdots \; (1)$

Sum of roots of the given quadratic equation: $\;$ $x_1 + x_2 = \dfrac{-b}{5}$ $\;\;\; \cdots \; (2)$

i.e. $\;$ $5x_1 + 5x_2 = \dfrac{-b}{5}$ $\;\;\; \cdots \; (2a)$

Solving equations $(1)$ and $(2a)$ simultaneously gives

$3x_2 = -b - 1$ $\implies$ $x_2 = \dfrac{-\left(b+1\right)}{3}$

Substituting the value of $x_2$ in equation $(2)$ gives

$x_1 = \dfrac{-b}{5} - x_2 = \dfrac{-b}{5} + \dfrac{b+1}{3}$ $\implies$ $x_1 = \dfrac{2b + 5}{15}$

Product of roots of the given quadratic equation: $\;$ $x_1 \cdot x_2 = \dfrac{-28}{5}$ $\;\;\; \cdots \; (3)$

Substituting the values of $x_1$ and $x_2$ in equation $(3)$ gives

$- \left(\dfrac{2b + 5}{15}\right) \left(\dfrac{b+1}{3}\right) = \dfrac{-28}{5}$

i.e. $\;$ $2b^2 + 7b + 5 = 252$

i.e. $\;$ $2b^2 + 7b - 247 = 0$

i.e. $\;$ $\left(b + 13\right) \left(2b - 19\right) = 0$

i.e. $\;$ $b = -13$ $\;$ or $\;$ $b = \dfrac{19}{2}$

$\because \;$ $b$ is an integer, $b = -13$ is the only acceptable solution.

Algebra - Equations and Inequations

For what value of $\;$ $a$ $\;$ is the difference between the roots of the equation $\left(a - 2\right)x^2 - \left(a - 4\right)x - 2 = 0$ equal to $3$?

Given quadratic equation: $\;\;\;$ $\left(a - 2\right)x^2 - \left(a - 4\right)x - 2 = 0$

Relation between the roots $\alpha$ and $\beta$ $\left(\alpha > \beta\right)$ of the given quadratic equation is

$\alpha - \beta = 3$ $\;\;\; \cdots \; (1)$

Sum of roots of the given quadratic equation: $\;$ $\alpha + \beta = \dfrac{a - 4}{a - 2}$ $\;\;\; \cdots \; (2)$

Adding equations $(1)$ and $(2)$ simultaneously gives

$2 \alpha = 3 + \dfrac{a - 4}{a - 2}$

i.e. $\;$ $2 \alpha = \dfrac{3a - 6 + a - 4}{a - 2}$

i.e. $\;$ $2 \alpha = \dfrac{4a - 10}{a - 2}$

i.e. $\;$ $\alpha = \dfrac{2a - 5}{a - 2}$

Substituting the value of $\alpha$ in equation $(2)$ gives

$\beta = \dfrac{a - 4}{a - 2} - \alpha = \dfrac{a - 4}{a - 2} - \dfrac{2a - 5}{a - 2}$

i.e. $\;$ $\beta = \dfrac{1 - a}{a - 2}$

Product of roots of the given quadratic equation: $\;$ $\alpha \cdot \beta = \dfrac{-2}{a - 2}$ $\;\;\; \cdots \; (3)$

Substituting the values of $\alpha$ and $\beta$ in equation $(3)$ gives

$\left(\dfrac{2a - 5}{a - 2}\right) \left(\dfrac{1 - a}{a - 2}\right) = \dfrac{-2}{a - 2}$

i.e. $\;$ $\left(2a - 5\right) \left(1 - a\right) = -2 \left(a - 2\right)$

i.e. $2a - 2a^2 - 5 + 5a = -2a + 4$

i.e. $\;$ $2a^2 - 9a + 9 = 0$

i.e. $\;$ $\left(a - 3\right) \left(2a - 3\right) = 0$

i.e. $\;$ $a = 3$ $\;$ or $\;$ $a = \dfrac{3}{2}$

Algebra - Equations and Inequations

The roots $x_1$ and $x_2$ of the equation $x^2 + px + 12 = 0$ are such that $x_2 - x_1 = 1$. Find $p$.

Given quadratic equation: $\;\;\;$ $x^2 + px + 12 = 0$

Relation between the roots $x_1$ and $x_2$ of the given quadratic equation is

$x_2 - x_1 = 1$ $\;\;\; \cdots \; (1)$

Sum of roots of the given quadratic equation: $\;$ $x_1 + x_2 = -p$ $\;\;\; \cdots \; (2)$

Adding equations $(1)$ and $(2)$ simultaneously gives

$2 x_2 = 1 - p$ $\implies$ $x_2 = \dfrac{1-p}{2}$

Substituting the value of $x_2$ in equation $(1)$ gives

$x_1 = x_2 - 1 = \dfrac{1 - p}{2} - 1 = - \left(\dfrac{1 + p}{2}\right)$

Product of roots of the given quadratic equation: $\;$ $x_1 \cdot x_2 = 12$ $\;\;\; \cdots \; (3)$

Substituting the values of $x_1$ and $x_2$ in equation $(3)$ gives

$- \left(\dfrac{1 - p}{2}\right) \left(\dfrac{1 + p}{2}\right) = 12$

i.e. $\;$ $1 - p^2 = - 48$

i.e. $p^2 = 49$

i.e. $\;$ $p = \pm 7$

Algebra - Equations and Inequations

Find $\;$ $a$ $\;$ such that one of the roots of the equation $\;$ $x^2 - \dfrac{15}{4} x + a = 0$ $\;$ is the square of the other.

Given quadratic equation: $\;\;\;$ $x^2 - \dfrac{15}{4} x + a = 0$ $\;\;\; \cdots \; (1)$

Given: $\;$ One root of equation $(1)$ is the square of the other.

Let the roots of $(1)$ be $\alpha, \; \alpha^2$.

Sum of roots of equation $(1)$ $= \alpha + \alpha^2 = \dfrac{- \left(-15/4\right)}{1} = \dfrac{15}{4}$

i.e. $\;$ $\alpha^2 + \alpha - \dfrac{15}{4} = 0$

i.e. $\;$ $4 \alpha^2 + 4 \alpha - 15 = 0$

i.e. $\;$ $\left(2 \alpha + 5\right) \left(2 \alpha - 3\right) = 0$

i.e. $\;$ $\alpha = \dfrac{-5}{2}$, $\;\;$ or $\;\;$ $\alpha = \dfrac{3}{2}$

Product of roots of equation $(1)$ $= \alpha \times \alpha^2 = \alpha^3 = \dfrac{a}{1} = a$ $\;\;\; \cdots \; (2)$

Substituting $\alpha = \dfrac{-5}{2}$ in equation $(2)$ gives

$a = \left(\dfrac{-5}{2}\right)^3 = \dfrac{-125}{8}$

Substituting $\alpha = \dfrac{3}{2}$ in equation $(2)$ gives

$a = \left(\dfrac{3}{2}\right)^3 = \dfrac{27}{8}$

Algebra - Equations and Inequations

Find the value of $\;$ $a$ $\;$ for which one root of the equation $\;$ $x^2 + \left(2a-1\right)x + a^2 + 2 = 0$ $\;$ is twice as large as the other.

Given quadratic equation: $\;\;\;$ $x^2 + \left(2a-1\right)x + a^2 + 2 = 0$ $\;\;\; \cdots \; (1)$

Given: $\;$ One root of equation $(1)$ is twice as large as the other.

Let the roots of $(1)$ be $\alpha, \; 2 \alpha$.

Sum of roots of equation $(1)$ $= \alpha + 2 \alpha = 3 \alpha = \dfrac{- \left(2a - 1\right)}{1} = 1 - 2a$

$\implies$ $\alpha = \dfrac{1 - 2a}{3}$ $\;\;\; \cdots \; (2)$

Product of roots of equation $(1)$ $= \alpha \times 2 \alpha = 2 \alpha^2 = \dfrac{a^2 + 2}{1}$

i.e. $\;$ $2 \alpha^2 = a^2 + 2$ $\;\;\; \cdots \; (3)$

In view of equation $(2)$, equation $(3)$ becomes

$2 \times \left(\dfrac{1 - 2a}{3}\right)^2 = a^2 + 2$

i.e. $\;$ $\dfrac{2 \left(1 - 4a + 4a^2\right)}{9} = a^2 + 2$

i.e. $\;$ $2 - 8a + 8a^2 = 9a^2 + 18$

i.e. $\;$ $a^2 + 8a + 16 = 0$

i.e. $\;$ $\left(a + 4\right)^2 = 0$

$\implies$ $a = -4$

Algebra - Equations and Inequations

For what values of $a$ is the ratio of the roots of the equation $ax^2 - \left(a + 3\right)x + 3 = 0$ equal to $1.5$?

Given quadratic equation: $\;\;\;$ $ax^2 - \left(a + 3\right)x + 3 = 0$ $\;\;\; \cdots \; (1)$

Comparing equation $(1)$ with the standard quadratic equation $\;$ $Ax^2 + Bx + C = 0$ $\;$ gives

$A = a, \; B = - \left(a + 3\right), \; C = 3$

Roots of equation $(1)$ are $x = \dfrac{a + 3 \pm \sqrt{\left[-\left(a + 3\right)\right]^2 - 4 \times a \times 3}}{2 \times a}$

i.e. $\;$ $x = \dfrac{a + 3 \pm \sqrt{a^2 + 6a + 9 - 12a}}{2a}$

i.e. $\;$ $x = \dfrac{a + 3 \pm \sqrt{a^2 - 6a + 9}}{2a}$

i.e. $\;$ $x = \dfrac{a + 3 \pm \sqrt{\left(a - 3\right)^2}}{2a}$

i.e. $\;$ $x = \dfrac{a + 3 \pm \left(a - 3\right)}{2a}$

$\therefore \;$ The roots are $\;\;$ $x_1 = \dfrac{a + 3 + a - 3}{2a} = 1$, $\;$ $x_2 = \dfrac{a + 3 - a + 3}{2a} = \dfrac{3}{a}$

As per question, ratio of roots $= 1.5$

i.e. $\;$ $\dfrac{x_1}{x_2} = \dfrac{1}{3/a} = 1.5$ $\implies$ $a = 4.5$

or $\;$ $\dfrac{x_2}{x_1} = 1.5 = \dfrac{3/a}{1} = 1.5$ $\implies$ $a = 2$

Algebra - Equations and Inequations

For what values of $a$ is the sum of the roots of the equation $x^2 + \left(2-a-a^2\right)x - a^2 = 0$ equal to zero?

Given quadratic equation: $\;\;\;$ $x^2 + \left(2-a-a^2\right)x - a^2 = 0$ $\;\;\; \cdots \; (1)$

Let the roots of $(1)$ be $\alpha$ and $\beta$

Sum of roots $= \alpha + \beta = \dfrac{-\left(2 - a - a^2\right)}{1} = a^2 + a - 2$

as per question, $\;$ $\alpha + \beta = 0$

i.e. $\;$ $a^2 + a - 2 = 0$

i.e. $\;$ $\left(a + 2\right) \left(a - 1\right)= 0$

i.e. $\;$ $a = -2$ $\;$ or $\;$ $a = 1$

Algebra - Equations and Inequations

Find the least integral value of $k$ for which the equation $x^2 - 2 \left(k + 2\right)x + 12 + k^2 = 0$ has two different real roots.

Given quadratic equation: $\;\;\;$ $x^2 - 2 \left(k + 2\right) x + 12 + k^2 = 0$ $\;\;\; \cdots \; (1)$

Equation $(1)$ has two different real roots when its discriminant $\Delta > 0$

i.e. $\;$ $\Delta = \left[-2 \left(k + 2\right)\right]^2 - 4 \times 1 \times \left(12 + k^2\right) > 0$

i.e. $\;$ $4 \left(k^2 + 4k + 4\right) - 48 - 4k^2 >0$

i.e. $\;$ $4k^2 + 16k + 16 - 48 - 4k^2 > 0$

i.e. $\;$ $16 k > 32$

i.e. $\;$ $k > 2$

$\therefore \;$ The least value of $k$ for which equation $(1)$ has two different real roots is $k = 3$

Algebra - Equations and Inequations

For what values of $k$ is the inequality $x^2 - \left(k-3\right) x - k + 6 > 0$ valid for all real $x$?

Consider the quadratic equation $\;\;$ $x^2 - \left(k-3\right)x - k + 6 = 0$ $\;\;\; \cdots \; (1)$

Equation $(1)$ will have real roots when its discriminant

$\Delta = \left[- \left(k - 3\right)\right]^2 - 4 \times 1 \times \left(6 - k\right) > 0$

i.e. $\;$ $k^2 - 6k + 9 - 24 + 4k > 0$

i.e. $\;$ $k^2 - 2k - 15 > 0$

i.e. $\;$ $\left(k + 3\right) \left(k - 5\right) > 0$

i.e. $\;$ $k + 3 > 0$ $\;$ and $\;$ $k - 5 > 0$ $\;\;$ OR $\;\;$ $k + 3 < 0$ $\;$ and $\;$ $k - 5 < 0$

i.e. $\;$ $k > -3$ $\;$ and $\;$ $k > 5$ $\;\;$ OR $\;\;$ $k < -3$ $\;$ and $\;$ $k < 5$

i.e. $\;$ $k > 5$ $\;\;$ OR $\;\;$ $k < -3$

i.e. $\;$ $5 < k < -3$

$\therefore \;$ The given inequality is valid for all real $x$ when $k \in \left(-3, 5\right)$