Algebra - Equations and Inequations

Form a quadratic equation whose roots are the numbers $\dfrac{1}{10 - \sqrt{72}}$ and $\dfrac{1}{10 + 6 \sqrt{2}}$

Roots of the required quadratic equation are

$\alpha = \dfrac{1}{10 - \sqrt{72}} = \dfrac{1}{10 - 6 \sqrt{2}}$ $\;$ and $\;$ $\beta = \dfrac{1}{10 + 6 \sqrt{2}}$

Sum of roots $= \alpha + \beta$

$\begin{aligned} \alpha + \beta & = \dfrac{1}{10 - 6 \sqrt{2}} + \dfrac{1}{10 + 6 \sqrt{2}} \\\\ & = \dfrac{10 + 6 \sqrt{2} + 10 - 6 \sqrt{2}}{\left(10 + 6 \sqrt{2}\right) \left(10 - 6 \sqrt{2}\right)} \\\\ & = \dfrac{20}{100 - 72} \\\\ & = \dfrac{20}{28} \end{aligned}$

Product of roots $= \alpha \cdot \beta$

$\begin{aligned} \alpha \cdot \beta & = \left(\dfrac{1}{10 - 6 \sqrt{2}}\right) \times \left(\dfrac{1}{10 + 6 \sqrt{2}}\right) \\\\ & = \dfrac{1}{100 - 72} \\\\ & = \dfrac{1}{28} \end{aligned}$

Quadratic equation in terms of its roots is

$x^2 - \left(\text{sum of roots}\right) x + \text{product of roots} = 0$

$\therefore \;$ The required quadratic equation is

$x^2 - \dfrac{20}{28}x + \dfrac{1}{28} = 0$

i.e. $\;$ $28 x^2 - 20x + 1 = 0$

Algebra - Equations and Inequations

Find the values of the coefficient $a$ for which the curve $y = x^2 + ax + 25$ touches the $X$ axis.

When the given curve $\;\;$ $y = x^2 + ax + 25$ $\;$ touches the $X$ axis, its y coordinate $y = 0$

i.e. $\;$ $x^2 + ax + 25 = 0$ $\;\;\; \cdots \; (1)$

Since the given curve touches the $X$ axis (at one point)

$\implies$ equation $(1)$ should have two real equal roots

Now, for real equal roots of equation $(1)$,

its discriminant $= \Delta = a^2 - 4 \times 1 \times 25 = 0$

i.e. $\;$ $a^2 - 100 = 0$

i.e. $\;$ $a^2 = 100$

i.e. $\;$ $a = \pm 10$

Algebra - Equations and Inequations

For what values of $m$ does the equation $x^2 - x + m = 0$ possess no real roots?

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

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

$A = 1, \; B = -1, \; C = m$

Discriminant of equation $(1)$ is

$\Delta = B^2 - 4AC = \left(-1\right)^2 - 4 \times 1 \times m = 1 - 4m$

For no real roots, $\;$ $\Delta < 0$

i.e. $\;$ $1 - 4m < 0$

i.e. $\;$ $1 < 4m$

i.e. $\;$ $\dfrac{1}{4} < m$ $\;\;$ i.e. $\;\;$ $m > \dfrac{1}{4}$

$\therefore \;$ The given quadratic equation has no real roots $\;$ $\forall \; m \in \left(\dfrac{1}{4}, + \infty\right)$

Algebra - Equations and Inequations

For what values of $a$ does the equation $9x^2 - 2x + a = 6 - ax$ possess equal roots?

Given quadratic equation: $\;$ $9x^2 -2x+a = 6 - ax$

i.e. $\;$ $9x^2 + \left(a-2\right)x + a-6 = 0$ $\;\;\; \cdots \; (1)$

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

$A = 9, \; B = a-2, \; C = a-6$

Discriminant of equation $(1)$ is

$\Delta = B^2 - 4AC = \left(a-2\right)^2 - 4 \times 9 \times \left(a - 6\right)$

i.e. $\;$ $\Delta = a^2 - 4a + 4 - 36a + 216$

i.e. $\;$ $\Delta = a^2 - 40a + 220$

For equal roots, $\;$ $\Delta = 0$

i.e. $\;$ $a^2 - 40a + 220 = 0$

i.e. $\;$ $a = \dfrac{40 \pm \sqrt{1600 - 4 \times 1 \times 220}}{2}$

i.e. $\;$ $a = \dfrac{40 \pm \sqrt{1600-880}}{2}$

i.e. $\;$ $a = \dfrac{40 \pm \sqrt{720}}{2} = \dfrac{40 \pm 12 \sqrt{5}}{2} = 20 \pm 6 \sqrt{5}$

Algebra - Equations and Inequations

Solve for $x$: $\;\;\;$ $\left|5 - 2x\right| < 1$

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

Now, $\;$ $\left|5 - 2x\right| = 5 - 2x$ $\;$ (when $\;$ $5 - 2x > 0$), $\;$ equation $(1)$ becomes

$5 - 2x < 1$

i.e. $\;$ $4 < 2x$ $\implies$ $2 < x$ $\;\;\; \cdots \; (2a)$

$\left|5 - 2x\right| = -\left(5 - 2x\right)$ $\;$ (when $\;$ $5 - 2x < 0$), $\;$ equation $(1)$ becomes

$-5 + 2x < 1$

i.e. $\;$ $2x < 6$ $\implies$ $x < 3$ $\;\;\; \cdots \; (2b)$

From $(2a)$ and $(2b)$, solution of inequation $(1)$ is $\;$ $2 < x < 3$

Algebra - Equations and Inequations

Solve for $x$: $\;\;\;$ $\left|x+2\right| = 2 \left(3-x\right)$

Given problem: $\;\;\;$ $\left|x+2\right| = 2 \left(3-x\right)$ $\;\;\; \cdots \; (1)$

$\left|x+2\right| = x + 2$ $\;$ when $\;$ $x + 2 > 0$ $\;$ i.e. $\;$ $x > -2$

When $\;$ $\left|x+2\right| = x + 2$, $\;$ equation $(1)$ becomes

$x + 2 = 2 \left(3-x\right)$

i.e. $\;$ $x + 2 = 6 - 2x$

i.e. $\;$ $3x = 4$ $\implies$ $x = \dfrac{4}{3}$

$\because \;$ $x > -2$ $\;$ and $\;$ $\dfrac{4}{3} > 2$ $\implies$ $x = \dfrac{4}{3}$ $\;$ is a valid solution.

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

When $\;$ $\left|x+2\right| = - \left(x + 2\right)$, $\;$ equation $(1)$ becomes

$-x - 2 = 2 \left(3-x\right)$

i.e. $\;$ $-x - 2 = 6 - 2x$

i.e. $\;$ $x = 8$

$\because \;$ $x < -2$ $\;$ and $\;$ $8 \nless -2$ $\implies$ $x = 8$ $\;$ is not a valid solution.

$\therefore \;$ Solution of equation $(1)$ is $x = \dfrac{4}{3}$

Algebra - Binomial Theorem

Find $P^n_2$ if the fifth term of the expansion of $\left(\sqrt[3]{x} + \dfrac{1}{x}\right)^n$ does not depend on $x$.

Fifth term of the expansion of $\left(\sqrt[3]{x} + \dfrac{1}{x}\right)^n$ is

$T_5 = T_{4+1} = C^n_4 \; \left(x^{\frac{1}{3}}\right)^{n-4} \; \left(x^{-1}\right)^4$

i.e. $\;$ $T_5 = C^n_4 \; x^{\left(\frac{n}{3} - \frac{4}{3} - 4\right)}$

i.e. $\;$ $T_5 = C^n_4 \; x^{\left(\frac{n-16}{3}\right)}$

As per question, the fifth term in the binomial expansion is independent of $x$.

$\implies$ $\dfrac{n-16}{3} = 0$ $\implies$ $n = 16$

Now, $\;$ $P^n_2 = P^{16}_2 = \dfrac{16!}{\left(16 - 2\right)!} = \dfrac{16 \times 15 \times 14!}{14!} = 240$

Algebra - Binomial Theorem

The sum of the coefficients in the first three terms of the expansion of $\left(x^2 - \dfrac{2}{x}\right)^m$ is equal to $97$. Find the term of the expansion containing $x^4$.

In the expansion of $\left(x^2 - \dfrac{2}{x}\right)^m$,

first term $= T_1 = C^m_0 \; \left(x^2\right)^m = C^m_0 \times x^{2m}$

second term $= T_2 = C^m_1 \; \left(x^2\right)^{m-1} \; \left(\dfrac{-2}{x}\right) = -2 C^m_1 \times x^{2m-3}$

third term $= T_3 = C^m_2 \; \left(x^2\right)^{m-2} \; \left(\dfrac{-2}{x}\right)^2 = 4 \; C^m_2 \times x^{2m-6}$

$\therefore \;$ Coefficients in the first three terms of the expansion of $\left(x^2 - \dfrac{2}{x}\right)^m$ are $\;$ $C^m_0, \; -2 C^m_1, \; 4C^m_2$.

As per question, $\;\;\;$ $C^m_0 - 2 C^m_1 + 4 C^m_2 = 97$

i.e. $\;$ $1 - \dfrac{2 \times m!}{1! \left(m-1\right)!} + \dfrac{4 \times m!}{2! \left(m-2\right)!} = 97$

i.e. $\;$ $-2m + \dfrac{4m \left(m-1\right)}{2} = 96$

i.e. $\;$ $2m^2 - 4m = 96$

i.e. $\;$ $m^2 - 2m - 48 = 0$

i.e. $\;$ $\left(m-8\right) \left(m+6\right) = 0$

i.e. $\;$ $m = 8$ $\;$ or $\;$ $m = -6$

$\because \;$ $m$ cannot be negative $\implies$ $m = 8$

Let $\left(r+1\right)^{th}$ term in the expansion of $\left(x^2 - \dfrac{2}{x}\right)^m = \left(x^2 - \dfrac{2}{x}\right)^8$ $\;$ be the term of the expansion containing $x^4$.

Now, $\;$ $\left(r+1\right)^{th}$ term $= T_{r+1} = C^8_r \; \left(x^2\right)^{8-r} \; \left(\dfrac{-2}{x}\right)^r$

i.e. $\;$ $T_{r+1} = \left(-2\right)^r \; C^8_r \; x^{16-3r}$

$\therefore \;$ As per question, $\;$ $16 - 3r = 4$

i.e. $\;$ $3r = 12$ $\implies$ $r = 4$

$\therefore \;$ The term containing $x^4$ in the expansion of $\left(x^2 - \dfrac{2}{x}\right)^8$ is

$T_{4+1} = T_5 = \left(-2\right)^4 \; C^8_4 \; x^4$

i.e. $\;$ $T_5 = 16 \times \dfrac{8!}{4! \left(8-4\right)!} \times x^4$

i.e. $\;$ $T_5 = \dfrac{16 \times 8 \times 7 \times 6 \times 5 \times x^4}{4 \times 3 \times 2 \times 1} = 1120 x^4$

Algebra - Binomial Theorem

Find the term of the expansion of $\left(\sqrt[3]{x} - \dfrac{1}{\sqrt{x}}\right)^{15}$ which does not contain $x$.

$\left(k + 1\right)^{th}$ term in the expansion of $\left(a + b\right)^n$ is \;\; $T_{k+1} = C^n_k \; a^{n-k} \; b^k$

Let the $\left(k + 1\right)^{th}$ term in the expansion of $\left(\sqrt[3]{x} - \dfrac{1}{\sqrt{x}}\right)^{15} = \left(x^{\frac{1}{3}} - x^{\frac{-1}{2}}\right)^{15}$ be independent of $x$.

Now, $\;$ $T_{k+1} = C^{15}_k \; \left(x^{\frac{1}{3}}\right)^{15 - k} \; \left(x^{\frac{-1}{2}}\right)^{k} = C^{15}_k \; x^{5- \frac{k}{3} - \frac{k}{2}}$

Since this term is independent of $x$ $\implies$ $5 - \dfrac{k}{3} - \dfrac{k}{2} = 0$

i.e. $\;$ $5 = \dfrac{5k}{6}$ $\implies$ $k = 6$

i.e. $7^{th}$ term is independent of $x$.

$\therefore \;$ The term independent of $x$ in the expansion of $\left(\sqrt[3]{x} - \dfrac{1}{\sqrt{x}}\right)^{15}$ is

$T_{7} = T_{6+1} = C^{15}_{6} = \dfrac{15!}{6! \left(15 - 6\right)!} = \dfrac{15!}{6! \times 9!} = 5005$

Algebra - Binomial Theorem

Determine the ordinal number of the term of the expansion of $\left(\dfrac{3}{4} \sqrt[3]{a^2} + \dfrac{2}{3} \sqrt{a}\right)^{12}$ which contains $a^7$.

In the expansion of $\left(x+y\right)^n$, $\;$ $\left(r+1\right)^{th}$ term $= T_{r+1} = C^n_r \; x^{n-r} \; y^r$

$\left(r+1\right)^{th}$ term in the expansion of $\left(\dfrac{3}{4} \sqrt[3]{a^2} + \dfrac{2}{3} \sqrt{a}\right)^{12} = \left(\dfrac{3}{4} a^{\frac{2}{3}} + \dfrac{2}{3} a^{\frac{1}{2}}\right)^{12}$ is

$T_{r+1} = C^{12}_r \; \left(\dfrac{3}{4} a^{\frac{2}{3}}\right)^{12-r} \; \left(\dfrac{2}{3} a^{\frac{1}{2}}\right)^r$

i.e. $\;$ $T_{r+1} = C^{12}_r \times \dfrac{3^{12-r} \; a^{\frac{24-2r}{3}}}{2^{24-2r}} \times \dfrac{2^r \; a^{\frac{r}{2}}}{3^r}$

i.e. $\;$ $T_{r+1} = C^{12}_r \times \dfrac{3^{12-2r} \; a^{\frac{24-2r}{3} + \frac{r}{2}}}{2^{24-3r}}$

Let the $\left(r+1\right)^{th}$ term in the expansion contain $a^7$.

Then,

$\dfrac{24-2r}{3} + \dfrac{r}{2} = 7$

i.e. $\;$ $48 - 4r + 3r = 42$

i.e. $\;$ $r = 6$

$\therefore \;$ The ordinal number of the term in the expansion of $\left(\dfrac{3}{4} \sqrt[3]{a^2} + \dfrac{2}{3} \sqrt{a}\right)^{12}$ which contains $a^7$ is $6$.

Algebra - Binomial Theorem

Find $x$ in the binomial expansion $\left(\sqrt[3]{2} + \dfrac{1}{\sqrt[3]{3}}\right)^x$ if the ratio of the seventh term from the beginning of the binomial expansion to the seventh term from its end is $\dfrac{1}{6}$.

In the expansion of $\left(x+a\right)^n$

$\left(r+1\right)^{th}$ term from the beginning $\;$ $= T_{r+1} = C^n_r \; x^{n-r} \; a^r$

$r^{th}$ term from the end in the expansion of $\left(x+a\right)^n$ is the $\left(n - r + 2\right)^{th}$ term from the beginning

i.e. $\;$ $r^{th}$ term from the end $\;$ $= T_{n-r+2}$

Now, in the expansion of $\left(\sqrt[3]{2} + \dfrac{1}{\sqrt[3]{3}}\right)^x$

$7^{th}$ term from the beginning $= T_7 = T_{6+1} = C^x_6 \; \left(\sqrt[3]{2}\right)^{x-6} \; \left(\dfrac{1}{\sqrt[3]{3}}\right)^6$

$7^{th}$ term from the end $= T_{x-7+2} = T_{x-5} = T_{x-6+1} = C^x_{x-6} \; \left(\sqrt[3]{2}\right)^{x-x+6} \; \left(\dfrac{1}{\sqrt[3]{3}}\right)^{x-6}$

i.e. $\;$ $T_{x-5} = C^x_{x-6} \; \left(\sqrt[3]{2}\right)^6 \; \left(\dfrac{1}{\sqrt[3]{3}}\right)^{x-6}$

As per question,

$\dfrac{T_7}{T_{x-5}} = \dfrac{1}{6}$

i.e. $\;$ $\dfrac{C^x_6 \; \left(\sqrt[3]{2}\right)^{x-6} \; \left(\dfrac{1}{\sqrt[3]{3}}\right)^6}{C^x_{x-6} \; \left(\sqrt[3]{2}\right)^6 \; \left(\dfrac{1}{\sqrt[3]{3}}\right)^{x-6}} = \dfrac{1}{6}$

Since $\;$ $C^n_r = C^n_{n-r}$ $\implies$ $C^x_6 = c^x_{x-6}$

$\therefore \;$ We have

$\left(\sqrt[3]{2}\right)^{x-12} \times \left(\dfrac{1}{\sqrt[3]{3}}\right)^{12-x} = \dfrac{1}{6}$

i.e. $\;$ $\dfrac{2^{\frac{x}{3}} \times 2^{\frac{-12}{3}}}{3^{\frac{12}{3}} \times 3^{\frac{-x}{3}}} = \dfrac{1}{6}$

i.e. $\;$ $\dfrac{2^{\frac{x}{3}} \times 2^{-4}}{3^{4} \times 3^{\frac{-x}{3}}} = \dfrac{1}{6}$

i.e. $\;$ $\dfrac{2^{\frac{x}{3}} \times 3^{\frac{x}{3}}}{2^4 \times 3^4} = \dfrac{1}{2 \times 3}$

i.e. $\;$ $2^{\frac{x}{3}} \times 3^{\frac{x}{3}} = 2^3 \times 3^3$

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

$\implies$ $x = 9$

Algebra - Elements of Combinatorics

How many five-digit numbers, which do not contain identical digits, can be written by means of the digits $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$ and $9$?

Given digits: $\;$ $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$ and $9$ $\;$ i.e. $\;$ $9$ digits

Since the digits are not repeated,

Ten thousand's place can be selected in $9$ ways

Thousand's place can be selected in $8$ ways

Hundred's place can be selected in $7$ ways

Ten's place can be selected in $6$ ways

Unit's place can be selected in $5$ ways

$\therefore \;$ Number of five-digit numbers that can be formed with the given digits so that the digits are not repeated

$= 9 \times 8 \times 7 \times 6 \times 5 = 15120$ numbers

Algebra - Binomial Theorem

Find the third term of the expansion of $\left(z^2 + \dfrac{1}{z} \sqrt[3]{z}\right)^n$ if the sum of all the binomial coefficients is equal to $2048$.

Sum of all binomial coefficients is

$C^n_0 + C^n_1 + C^n_2 + \cdots + C^n_n = 2^n$

Then, as per question,

$2^n = 2048$

i.e. $\;$ $2^n = 2^{11}$ $\implies$ $n = 11$

$\therefore \;$ the problem is to find the third term in the expansion of

$\begin{aligned} \left(z^2 + \dfrac{1}{z} \sqrt[3]{z}\right)^n & = \left(z^2 + z^{\frac{1}{3} - 1}\right)^{11} \\\\ & = \left(z^2 + z^{\frac{-2}{3}}\right)^{11} \end{aligned}$

$\therefore \;$ The required third term is

$\begin{aligned} T_3 = T_{2+1} & = C^{11}_{2} \; \left(z^2\right)^{11 - 2} \; \left(z^{\frac{-2}{3}}\right)^2 \\\\ & = \dfrac{11!}{2! \times \left(11 - 2\right)!} \times \left(z^2\right)^9 \times z^{\frac{-4}{3}} \\\\ & = \dfrac{11 \times 10 \times 9!}{2 \times 9!} \times z^{18 - \frac{4}{3}} \\\\ & = 55 z^{\frac{50}{3}} \end{aligned}$

Algebra - Elements of Combinatorics

How many four-digit numbers which are divisible by $4$ can be formed from the digits $1$, $2$, $3$, $4$ and $5$ if the digits are repeated?

A number is divisible by $4$ if the digits in its unit's and ten's place are divisible by $4$.

Since the digits are repeated, the digits in the ten's and the unit's place can be

$1$ and $2$ $\;$ OR $\;$ $2$ and $4$ $\;$ OR $\;$ $3$ and $2$ $\;$ OR $\;$ $4$ and $4$ $\;$ OR $\;$ $5$ and $2$ $\;$ respectively

i.e. $\;$ The digits in the ten's and the unit's place can be selected in $5$ ways.

For each of these selections,

the hundred's place can be selected in $5$ ways

and the thousand's place can be selected in $5$ ways.

$\therefore \;$ Total number of four digit numbers that can be formed from the digits $1$, $2$, $3$, $4$ and $5$ (digits repeated) which are divisible by $4$ are

$5 \times 5 \times 5 = 125$ numbers

Algebra - Binomial Theorem

Find the second term of the binomial expansion of $\left(\sqrt[13]{a} + \dfrac{a}{\sqrt{a^{-1}}}\right)^m$ if $C^m_3 : C^m_2 = 4 : 1$.

Given: $\;\;$ $C^m_3 : C^m_2 = 4 : 1$

i.e. $\;$ $\dfrac{m!}{3! \left(m-3\right)!} : \dfrac{m!}{2! \left(m-2\right)!} = \dfrac{4}{1}$

i.e. $\;$ $\dfrac{m!}{3 \times 2! \left(m-3\right)!} \times \dfrac{2! \left(m-2\right) \left(m-3\right)!}{m!} = 4$

i.e. $\;$ $\dfrac{m-2}{3} = 4$

i.e. $\;$ $m - 2 = 12$ $\implies$ $m = 14$

$\therefore \;$ The problem is: $\;\;$ $\left(\sqrt[13]{a} + \dfrac{a}{\sqrt{a^{-1}}}\right)^m = \left(a^{\frac{1}{13}} + a^{1 + \frac{1}{2}}\right)^{14} = \left(a^{\frac{1}{13}} + a^{\frac{3}{2}}\right)^{14}$

$\therefore \;$ Second term in the expansion of $\left(a^{\frac{1}{13}} + a^{\frac{3}{2}}\right)^{14}$ is

$\begin{aligned} T_2 = T_{1+1} & = C^{14}_{1} \; \left(a^{\frac{1}{13}}\right)^{14-1} \; \left(a^{\frac{3}{2}}\right)^1 \\\\ & = 14 \times \left(a^{\frac{1}{13}}\right)^{13} \times a^{\frac{3}{2}} \\\\ & = 14 \; a^{1+\frac{3}{2}} \\\\ & = 14a^{\frac{5}{2}} \end{aligned}$

Algebra - Elements of Combinatorics

How many six-digit numbers can be formed from the digits $1$, $2$, $3$, $4$, $5$, $6$ and $7$ so that the digits do not repeat and the terminal digits should be even.

Given: $\;$ Three even digits -- $2$, $4$ and $6$

There are two terminal digits in a six digit number.

$2$ terminal digits from given $3$ even digits can be selected in

$P^3_2 = \dfrac{3!}{\left(3-2\right)!} = 3! = 6$ ways

Since the digits are not repeated, the remaining $4$ digits of the number can be selected from the remaining given $5$ digits in

$P^5_4 = 5! = 120$ ways

$\therefore \;$ Total number of six-digit numbers that can be formed from the given digits so that the digits are not repeated and the terminal digits are even are

$6 \times 120 = 720$ numbers

Algebra - Binomial Theorem

Find the term of the expansion of $\left(\sqrt[3]{x^{-2}} + x\right)^7$ containing $x$ in the second power.

$\left(r+1\right)^{th}$ term in the expansion of $\left(a+b\right)^n$ is $\;\;$ $T_{r+1} = C^n_r \; a^{n-r} \; b^r$

Here, $\;$ $n = 7, \; a = \sqrt[3]{x^{-2}} = x^{\frac{-2}{3}}, \; b = x$

Let the $\left(r+1\right)^{th}$ term in the expansion of $\left(\sqrt[3]{x^{-2}} + x\right)^7$ contain $x$ in the second power.

Now, $\left(r+1\right)^{th}$ term is

$\begin{aligned} T_{r+1} & = C^7_r \; \left(x^{\frac{-2}{3}}\right)^{7-r} \; x^r \\\\ & = C^7_r \; x^{\frac{-14}{3} + \frac{2r}{3} + r} \\\\ & = C^7_r \; x^{\frac{5r-14}{3}} \end{aligned}$

$\because \;$ the $\left(r+1\right)^{th}$ term contains $x$ in the second power,

$\implies$ $\dfrac{5r - 14}{3} = 2$

i.e. $\;$ $5r - 14 = 6$

i.e. $\;$ $5r = 20$

$\implies$ $r = 4$

$\therefore \;$ The required term is

$\begin{aligned} T_{4+1} = T_5 & = C^7_4 \; x^{\frac{20 - 14}{3}} \\\\ & = \dfrac{7!}{4! \times 3!} x^2 \\\\ & = \dfrac{7 \times 6 \times 5 \times 4!}{4! \times 6} x^2 \\\\ & = 35 x^2 \end{aligned}$

Algebra - Elements of Combinatorics

Five boys and three girls are sitting in a row of eight seats. In how many ways can they be seated so that not all the girls sit side by side?

Consider the three girls as $1$ group.

Then there are $5$ boys AND $1$ group of three girls, i.e. $6$ people.

These can be seated amongst themselves in $6! = 720$ ways.

Amongst themselves, the $3$ girls can be seated in $3! = 6$ ways

$\therefore \;$ Number of ways in which $5$ boys AND $1$ group of three girls can be seated

$= 3! \times 6! = 6 \times 720 = 4320$ ways.

Now, $3$ girls and $5$ boys, i.e. $8$ people can be seated amongst themselves in $8! = 40320$ ways.

$\therefore \;$ Number of ways in which $3$ girls and $5$ boys can be seated so that not all the girls sit next to one another

$= 40320 - 4320 = 36000$ ways

Algebra - Binomial Theorem

Find the term in the expansion of $\left(x + \dfrac{1}{x}\right)^8$ which does not contain $x$.

By binomial theorem, $\;$ $\left(a + b\right)^n = C^n_0 \; a^n + C^n_1 \; a^{n-1} \; b + \cdots + C^n_k \; a^{n-k} \; b^k + \cdots + C^n_n \; b^n$

$\left(k + 1\right)^{th}$ term in the expansion of $\left(a + b\right)^n$ is \;\; $T_{k+1} = C^n_k \; a^{n-k} \; b^k$

Let the $\left(k + 1\right)^{th}$ term in the expansion of $\left(x + \dfrac{1}{x}\right)^8$ be independent of $x$.

Now, $\;$ $T_{k+1} = C^8_k \; x^{8-k} \; \left(\dfrac{1}{x}\right)^{k} = C^8_k \; x^{8-k-k} = C^8_k \; x^{8-2k}$

Since this term is independent of $x$ $\implies$ $8 - 2k = 0$ $\implies$ $k = 4$

i.e. $5^{th}$ term is independent of $x$.

$\therefore \;$ The term independent of $x$ in the expansion of $\left(x + \dfrac{1}{x}\right)^8$ is

$T_5 = T_{4+1} = C^8_4 = \dfrac{8!}{4! \left(8 - 4\right)!} = \dfrac{8 \times 7 \times 6 \times 5 \times 4!}{4! \times 4 \times 3 \times 2 \times 1} = 70$

Algebra - Elements of Combinatorics

In how many ways can $10$ identical presents be distributed among $6$ children so that each child gets at least one present.

Let $1$ present be given to each child. This can be done in $1$ way ($\because \;$ all presents are identical).

Then we are left with $4$ identical presents to be distributed among $6$ children.

This can be done as follows:

1. Give all $4$ presents to one child.
   
   This can be done in $\;$ $C^6_1 = \dfrac{6!}{1! \times 5!} = 6$ ways
   
   OR



2. Give $3$ presents to one child and $1$ present to another.
   
   This can be done in $\;$ $C^6_1 \times C^5_1 = 6 \times 5 = 30$ ways
   
   OR



3. Give $2$ presents to one child and $2$ to another child; i.e. select $2$ children from $6$ children.
   
   This can be done in $\;$ $C^6_2 = \dfrac{6!}{2! \; 4!} = \dfrac{6 \times 5 \times 4!}{2 \times 4!} = 15$ ways
   
   OR



4. Give $2$ presents to one child and $1$ present each to two children.
   
   This can be done in $\;$ $C^6_1 \times C^5_2 = 6 \times \dfrac{5!}{2! \times 3!} = 6 \times \dfrac{5 \times 4 \times 3!}{2 \times 3!} = 60$ ways
   
   OR



5. Give $1$ present to one child each; i.e. select $4$ children from $6$ children.
   
   This can be done in $\;$ $C^6_4 = \dfrac{6!}{4! 2!} = \dfrac{6 \times 5 \times 4!}{4! \times 2} = 15$ ways

$\therefore \;$ Total number of ways in which $10$ identical presents be distributed among $6$ children is

$= 1 \times \left(6 + 30 + 15 + 60 + 15\right) = 126$ ways

Algebra - Binomial Theorem

In the expansion of $\left(a \sqrt{a} + \dfrac{1}{a^4}\right)^n$, the coefficient in the second term exceeds by 44 the coefficient in the first term. Find $n$.

In the expansion of $\left(a \sqrt{a} + \dfrac{1}{a^4}\right)^n$,

coefficient in first term $= C^n_1$; $\;$ coefficient in second term $= C^n_2$

As per question,

$C^n_2 = C^n_1 + 44$

i.e. $\;$ $\dfrac{n!}{2!\left(n-2\right)!} = \dfrac{n!}{\left(n-1\right)!} + 44$

i.e. $\;$ $\dfrac{n \left(n-1\right) \left(n-2\right)!}{2\left(n-2\right)!} = \dfrac{n \left(n-1\right)!}{\left(n-1\right)!} + 44$

i.e. $\;$ $\dfrac{n \left(n-1\right)}{2} = n + 44$

i.e. $\;$ $n \left(\dfrac{n-1}{2} - 1\right) = 44$

i.e. $\;$ $\dfrac{n \left(n-3\right)}{2} = 44$

i.e. $\;$ $n^2 - 3n - 88 = 0$

i.e. $\;$ $\left(n-11\right)\left(n+8\right) = 0$

i.e. $\;$ $n = 11$ $\;$ or $\;$ $n = -8$

$\because \;$ $n$ cannot be negative as it is the total number of objects

$\therefore \;$ $n = 11$

Algebra - Elements of Combinatorics

In a $12$ storey house, $9$ people enter a lift cabin. It is known that they will leave the lift in groups of $2$, $3$ and $4$ people at different storeys. In how many ways can they do so if the lift does not stop at the second storey?

$9$ people can be grouped in groups of $2$, $3$ and $4$ people as follows:

$1$ group of two people $+$ $1$ group of three people $+$ $1$ group of four people $\;$ i.e. $\;$ $3$ groups

$0$ group of two people $+$ $3$ groups of three people $+$ $0$ group of four people $\;$ i.e. $\;$ $3$ groups

$3$ groups of two people $+$ $1$ group of three people $+$ $0$ group of four people $\;$ i.e. $\;$ $4$ groups

$\therefore \;$ Total number of possible groups $= 3 + 3 + 4 = 10$ groups

Now, $3$ groups can be selected from $10$ possible groups in $P^{10}_{3}$ ways.

$\therefore \;$ $9$ people entering the lift can leave in groups of $2$, $3$ and $4$ people in $P^{10}_{3}$ ways.

Algebra - Binomial Theorem

Find the middle term of the expansion of $\left(\sqrt{x} - \dfrac{1}{x}\right)^6$

There are $7$ terms in the expansion of $\left(\sqrt{x} - \dfrac{1}{x}\right)^6$.

The $4^{th}$ term will be the middle term.

Now, the $\left(r + 1\right)^{th}$ term $T_{r+1}$ in the expansion of $\left(p + q\right)^n$ is

$T_{r+1} = C^n_r \; p^{n-r} \; q^r$

$\therefore \;$ In the given problem,

Fourth term $= T_4 = T_{3+1} = C^{6}_{3} \; \left(\sqrt{x}\right)^{6-3} \; \left(\dfrac{-1}{x}\right)^3$

$\begin{aligned} i.e. \; T_4 & = \dfrac{6!}{3! \left(6-3\right)!} \times \left(\sqrt{x}\right)^3 \times \left(\dfrac{-1}{x^3}\right) \\\\ & = \dfrac{-6 \times 5 \times 4 \times 3!}{3! \times 6} \times x^{\frac{3}{2} - 3} \\\\ & = -20 x^{\frac{-3}{2}} \end{aligned}$

Algebra - Elements of Combinatorics

In how many ways can $5$ boys and $5$ girls be seated at a table so that no two boys and no two girls sit side by side?

$5$ boys can be seated at the table in $\;$ $P_5 = 5! = 120$ $\;$ ways

$5$ girls can be seated at the table in $\;$ $P_5 = 5! = 120$ $\;$ ways

Amongst themselves, the boys and girls can be seated in $\;$ $2$ $\;$ ways

$\therefore \;$ Total number of ways in which $5$ boys and $5$ girls can be seated at a table

$= 2 \times \left(P_5\right)^2 = 2 \times 120^2 = 28800$ ways

Algebra - Elements of Combinatorics

How many positive terms are there in the sequence $\left(x_n\right)$ if $x_n = \dfrac{195}{4 \times P_n} - \dfrac{P^{n+3}_{3}}{P_{n+1}}$, $\;\;\;$ $n \in N$

$\begin{aligned} x_n & = \dfrac{195}{4 \times P_n} - \dfrac{P^{n+3}_{3}}{P_{n+1}} \\\\ & = \dfrac{195}{4 \times n!} - \dfrac{\left(n+3\right)!}{\left(n+3-3\right)! \left(n+1\right)!} \\\\ & = \dfrac{195}{4 \times n!} - \dfrac{\left(n+3\right) \left(n+2\right) \left(n+1\right)!}{n! \left(n+1\right)!} \\\\ & = \dfrac{195}{4 \times n!} - \dfrac{\left(n+3\right) \left(n+2\right)}{n!} \\\\ & = \dfrac{195 - 4 \left(n^2 + 5n + 6\right)}{4 \times n!} \\\\ & = \dfrac{195 - 4n^2 - 20n - 24}{4 \times n!} \\\\ & = \dfrac{171 - 20n - 4n^2}{ 4 \times n!} \end{aligned}$

When $\;$ $n = 1$, $\;$ $x_n = \dfrac{171 - 20 - 4}{4 \times 1!} = \dfrac{147}{4 \times 1} > 0$

When $\;$ $n = 2$, $\;$ $x_n = \dfrac{171 - 40 - 16}{4 \times 2!} = \dfrac{115}{4 \times 2} > 0$

When $\;$ $n = 3$, $\;$ $x_n = \dfrac{171 - 60 - 36}{4 \times 3!} = \dfrac{75}{4 \times 6} > 0$

When $\;$ $n = 4$, $\;$ $x_n = \dfrac{171 - 80 - 64}{4 \times 4!} = \dfrac{27}{4 \times 24} > 0$

When $\;$ $n = 5$, $\;$ $x_n = \dfrac{171 - 100 - 100}{4 \times 5!} = \dfrac{-29}{4 \times 120} < 0$

$\therefore \;$ There are $4$ positive terms in the given sequence.

Algebra - Elements of Combinatorics

Solve the inequation: $\;$ $\dfrac{P^{n+1}_{4}}{C^{n-1}_{n-3}} > 14 \times P_3$, $\;$ $n \in N$

$\dfrac{P^{n+1}_{4}}{C^{n-1}_{n-3}} > 14 \times P_3$

i.e. $\;$ $\dfrac{\left(n + 1\right)!}{\left(n + 1 - 4\right)!} \div \dfrac{\left(n-1\right)!}{\left(n-3\right)! \left(n-1-n+3\right)!} > 14 \times 3!$

i.e. $\;$ $\dfrac{\left(n+1\right)!}{\left(n-3\right)!} \times \dfrac{\left(n-3\right)! \times 2!}{\left(n-1\right)!} > 14 \times 6$

i.e. $\;$ $\dfrac{2 \left(n+1\right) n \left(n-1\right)!}{\left(n-1\right)!} > 14 \times 6$

i.e. $\;$ $n \left(n+1\right) > 14 \times 3$

i.e. $\;$ $n^2 + n - 42 > 0$

i.e. $\;$ $\left(n+7\right) \left(n-6\right) > 0$

i.e. $\;$ $n+7 > 0$ $\;$ and $\;$ $n - 6 > 0$ $\;\;$ OR $\;\;$ $n + 7 < 0$ $\;$ and $\;$ $n - 6 < 0$

i.e. $\;$ $n > -7$ $\;$ and $\;$ $n > 6$ $\;\;$ OR $\;\;$ $n < -7$ $\;$ and $\;$ $n < 6$

i.e. $\;$ $n = \left\{-6, -5, -4, \cdots, 0, 1, 2, \cdots, 5, 6, 7, 8, \cdots\right\} \cap \left\{7, 8, 9, \cdots\right\}$

OR $\;\;$ $n = \left\{-8, -9, -10, \cdots\right\} \cap \left\{5, 4, 3, 2, 1, 0, -1, \cdots, -6, -7, -8, -9, \cdots\right\}$

i.e. $\;$ $n = \left\{7, 8, 9, \cdots\right\}$ $\;\;$ OR $\;\;$ $n = \left\{-8, -9, -10, \cdots\right\}$

$\because \;$ $n \in N$ $\implies$ $n = \left\{7, 8, 9, \cdots\right\}$

Algebra - Elements of Combinatorics

Solve the inequation: $\;$ $C^{x+1}_{x-1} > \dfrac{3}{2}$, $\;$ $x \in N$

$C^{x+1}_{x-1} > \dfrac{3}{2}$

i.e. $\;$ $\dfrac{\left(x+1\right)!}{\left(x-1\right)! \left(x+1-x+1\right)!} > \dfrac{3}{2}$

i.e. $\;$ $\dfrac{\left(x+1\right) x \left(x-1\right)!}{\left(x-1\right)! 2!} > \dfrac{3}{2}$

i.e. $\;$ $\dfrac{\left(x+1\right) x}{2} > \dfrac{3}{2}$

i.e. $\;$ $x^2 + x > 3$

i.e. $\;$ $x^2 + x - 3 > 0$ $\;\;\; \cdots \; (1)$

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

$x = \dfrac{-1 \pm \sqrt{1 - 4 \times 1 \times \left(-3\right)}}{2 \times 1} = \dfrac{-1 \pm \sqrt{13}}{2}$

i.e. $\;$ $x = \dfrac{-1 + \sqrt{13}}{2} = 1.302$ $\;$ or $\;$ $x = \dfrac{-1 - \sqrt{13}}{2}$

$\because \;$ $x \in N$, $\;$ the possible solution to $(1)$ is $\;$ $x \geq 2$

Algebra - Elements of Combinatorics

Solve the inequation: $\;$ $C^{x+1}_{x-1} < 21$, $\;$ $x \in N$

$C^{x+1}_{x-1} < 21$

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

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

i.e. $\;$ $\left(x + 1\right) x < 42$

i.e. $\;$ $x^2 + x - 42 < 0$

i.e. $\;$ $\left(x + 7\right) \left(x - 6\right) < 0$

i.e. $\;$ $x + 7 < 0$ $\;$ and $\;$ $x - 6 > 0$ $\;\;$ OR $\;\;$ $x + 7 > 0$ $\;$ and $\;$ $x - 6 < 0$

i.e. $\;$ $x < -7$ $\;$ and $\;$ $x > 6$ $\;\;$ OR $\;\;$ $x > -7$ $\;$ and $\;$ $x < 6$

$\implies$ $x = \left\{1, 2, 3, 4, 5\right\}$ $\;$ $\because \; x \in N$

Algebra - Elements of Combinatorics

Solve the inequation: $\;$ $C^{x-1}_{4} - C^{x-1}_{3} - \dfrac{5}{4} P^{x-2}_{2} < 0$, $\;$ $x \in N$

$C^{x-1}_{4} - C^{x-1}_{3} - \dfrac{5}{4} P^{x-2}_{2} < 0$

i.e. $\;$ $C^{x-1}_{4} - C^{x-1}_{3} < \dfrac{5}{4} P^{x-2}_{2}$

i.e. $\;$ $\dfrac{\left(x - 1\right)!}{4! \left(x - 1 - 4\right)!} - \dfrac{\left(x - 1\right)!}{3! \left(x - 1 - 3\right)!} < \dfrac{5 \times \left(x - 2\right)!}{4 \times \left(x - 2 - 2\right)!}$

i.e. $\;$ $\dfrac{\left(x - 1\right)!}{4! \left(x - 5\right)!} - \dfrac{\left(x - 1\right)!}{3! \left(x - 4\right)!} < \dfrac{5 \times \left(x - 2\right)!}{4 \times \left(x - 4\right)!}$

i.e. $\;$ $\dfrac{\left(x - 1\right) \left(x - 2\right)!}{24 \left(x - 5\right)!} - \dfrac{\left(x - 1\right) \left(x - 2\right)!}{6 \left(x - 4\right) \left(x - 5\right)!} < \dfrac{5 \times \left(x - 2\right)!}{4 \left(x - 4\right) \left(x - 5\right)!}$

i.e. $\;$ $\dfrac{x-1}{12} - \dfrac{x-1}{3 \left(x - 4\right)} < \dfrac{5}{2 \left(x - 4\right)}$

i.e. $\;$ $\dfrac{\left(x - 1\right) \left(x - 4\right) - 4 \left(x - 1\right)}{12 \left(x - 4\right)} < \dfrac{5}{2 \left(x - 4\right)}$

i.e. $\;$ $\dfrac{x^2 - 5x + 4 - 4x + 4}{6} < 5$

i.e. $\;$ $x^2 - 9x + 8 < 30$

i.e. $\;$ $x^2 - 9x - 22 < 0$

i.e. $\;$ $\left(x - 11\right) \left(x + 2\right) < 0$

$\implies$ $x - 11 <0$ $\;$ and $\;$ $x + 2 > 0$ $\; \;$ OR $\;\;$ $x - 11 > 0$ $\;$ and $\;$ $x + 2 < 0$

i.e. $\;$ $x < 11$ $\;$ and $\;$ $x > -2$ $\;\;$ OR $\;\;$ $x > 11$ $\;$ and $\;$ $x < -2$

$\because \;$ $x \in N$ $\implies$ $x = \left\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\right\}$

Now, as per the question, $x - 1$ elements are taken $4$ at a time

$\implies$ $x = \left\{5, 6, 7, 8, 9,10\right\}$

Algebra - Elements of Combinatorics

Solve the inequation: $\;$ $5 C^{n}_{3} < C^{n+2}_{4}$, $\;$ $n \in N$

$5 C^{n}_{3} < C^{n+2}_{4}$

i.e. $\;$ $\dfrac{5 \times n!}{3! \left(n - 3\right)!} < \dfrac{\left(n + 2\right)!}{4! \left(n + 2 - 4\right)!}$

i.e. $\;$ $\dfrac{5 \times n!}{3! \left(n - 3\right)!} < \dfrac{\left(n + 2\right)!}{4! \left(n - 2\right)!}$

i.e. $\;$ $\dfrac{5 \times n!}{3! \left(n - 3\right)!} < \dfrac{\left(n + 2\right) \left(n + 1\right) n!}{4 \times 3! \left(n - 2\right) \left(n - 3\right)!}$

i.e. $\;$ $5 < \dfrac{\left(n + 2\right) \left(n + 1\right)}{4 \left(n - 2\right)}$

i.e. $\;$ $20n - 40 < n^2 + 3n + 2$

i.e. $\;$ $0 < n^2 - 17n + 42$

i.e. $\;$ $n^2 - 17n + 42 > 0$

i.e. $\;$ $\left(x - 14\right) \left(x - 3\right) > 0$

i.e. $\;$ $x - 14 > 0$ $\;$ or $\;$ $x - 3 > 0$

i.e. $\;$ $x > 14$ $\;$ or $\;$ $x > 3$

$\implies$ $x > 14, \; n \in N$

Algebra - Elements of Combinatorics

Solve the inequation: $\;$ $C^{n}_{6} < C^{n}_{4}$, $\;$ $n \in N$

$C^{n}_{6} < C^{n}_{4}$

i.e. $\;$ $\dfrac{n!}{6! \left(n - 6\right)!} < \dfrac{n!}{4! \left(n - 4\right)!}$

i.e. $\;$ $4! \left(n - 4\right)! < 6! \left(n - 6\right)!$

i.e. $\;$ $4! \left(n - 4\right) \left(n - 5\right) \left(n - 6\right)! < 6 \times 5 \times 4! \left(n - 6\right)!$

i.e. $\;$ $\left(n - 4\right) \left(n - 5\right) < 30$

i.e. $\;$ $n^2 -9n + 20 < 30$

i.e. $\;$ $n^2 - 9n - 10 < 0$

i.e. $\;$ $\left(n - 10\right) \left(n + 1\right) < 0$

i.e. $\;$ $n - 10 < 0$ $\;$ or $\;$ $n + 1 < 0$

i.e. $\;$ $n < 10$ $\;$ or $n < -1$

$\because \;$ $n \in N$ $\implies$ $n = \left\{6, 7, 8, 9\right\}$ $\;$ as per the question, maximum number of $n$ elements taken at a time is $6$.

Algebra - Elements of Combinatorics

Solve the inequation: $\;$ $C^{18}_{m-2} > C^{18}_{m}$, $\;$ $m \in N$

$C^{18}_{m-2} > C^{18}_{m}$

i.e. $\;$ $\dfrac{18!}{\left(m - 2\right)! \left(18 - m + 2\right)!} > \dfrac{18!}{m! \left(18 - m\right)!}$

i.e. $\;$ $m! \left(18 - m\right)! > \left(m - 2\right)! \left(20 - m\right)!$

i.e. $\;$ $m \left(m - 1\right) \left(m - 2\right)! \left(18 - m\right)! > \left(m - 2\right)! \left(20 - m\right) \left(19 - m\right) \left(18 - m\right)!$

i.e. $\;$ $m \left(m - 1\right) > \left(20 - m\right) \left(19 - m\right)$

i.e. $\;$ $m^2 - m > 380 - 39m + m^2$

i.e. $\;$ $38m > 380$

i.e. $\;$ $m > 10$

$\because \;$ $m \in M$ $\implies$ $m = \left\{11, 12, 13, 14, 15, 16, 17, 18\right\}$ $\;$ as the total number of elements is $18$.

Algebra - Elements of Combinatorics

Solve the inequation: $\;$ $C^{13}_{m} < C^{13}_{m + 2}$, $\;$ $m \in N$

$C^{13}_{m} < C^{13}_{m + 2}$

i.e. $\;$ $\dfrac{13!}{m! \left(13 - m\right)!} < \dfrac{13!}{\left(m + 2\right)! \left(13 - m - 2\right)!}$

i.e. $\;$ $\left(m + 2\right)! \left(11 - m\right)! < m! \left(13 - m\right)!$

i.e. $\;$ $\left(m + 2\right) \left(m + 1\right) m! \left(11 - m\right)! < m! \left(13 - m\right) \left(12 - m\right) \left(11 - m\right)!$

i.e. $\;$ $\left(m + 2\right) \left(m + 1\right) < \left(13 - m\right) \left(12 - m\right)$

i.e. $\;$ $m^2 + 3m + 2 < 156 - 25m + m^2$

i.e. $\;$ $28m < 154$

i.e. $\;$ $m < \dfrac{154}{28}$

i.e. $\;$ $m < \dfrac{11}{2}$

$\because \;$ $m \in N$ $\implies$ $m = \left\{1, 2, 3, 4, 5\right\}$

Algebra - Elements of Combinatorics

Solve the equation: $\;$ $3 C^{x+1}_{2} + P_2 \cdot x = 4 P^x_2$, $\;$ $x \in N$

$3 C^{x+1}_{2} + P_2 \cdot x = 4 P^x_2$

i.e. $\;$ $\dfrac{3 \times\left(x + 1\right)!}{2! \left(x + 1 -2\right)!} + 2! \times x = \dfrac{4 \times x!}{\left(x - 2\right)!}$

i.e. $\;$ $\dfrac{3 \times \left(x + 1\right)!}{2 \times \left(x - 1\right)!} + 2x = \dfrac{4 x \left(x - 1\right) \left(x - 2\right)!}{\left(x - 2\right)!}$

i.e. $\;$ $\dfrac{3 \left(x + 1\right) x \left(x - 1\right)!}{2 \left(x - 1\right)!} + 2x = 4 x \left(x - 1\right)$

i.e. $\;$ $\dfrac{3 x \left(x + 1\right)}{2} + 2x = 4x \left(x - 1\right)$

i.e. $\;$ $3x + 3 + 4 = 8x - 8$

i.e. $\;$ $5x = 15$

$\implies$ $x = 3$