Question

Find the integral value of x, if $\left| \begin{matrix} {{x}^{2}} & x & 1 \\\ 0 & 2 & 1 \\\ 3 & 1 & 4 \\\ \end{matrix} \right|=28$

Answer 1

Hint: Any determinant of $3\times 3$ order $\left| \begin{matrix} {{x}_{1}} & {{y}_{1}} & {{z}_{1}} \\\ {{x}_{2}} & {{y}_{2}} & {{z}_{2}} \\\ {{x}_{3}} & {{y}_{3}} & {{z}_{3}} \\\ \end{matrix} \right|$ is given along column 1 as
${{x}_{1}}\left( {{y}_{2}}{{z}_{3}}-{{z}_{3}}{{y}_{2}} \right)-{{x}_{2}}\left( {{y}_{1}}{{z}_{3}}-{{y}_{3}}{{z}_{1}} \right)+{{x}_{3}}\left( {{y}_{1}}{{z}_{2}}-{{y}_{2}}{{z}_{1}} \right)$

Complete step-by-step answer:
Use the above rule for expansion of the given determinant and use the quadratic formula to get roots of the formed quadratic equation.
Given expression in the problem is
$\left| \begin{matrix} {{x}^{2}} & x & 1 \\\ 0 & 2 & 1 \\\ 3 & 1 & 4 \\\ \end{matrix} \right|=28.............................\left( i \right)$
As we know the expansion of any determinant $\left| \begin{matrix} {{x}_{1}} & {{y}_{1}} & {{z}_{1}} \\\ {{x}_{2}} & {{y}_{2}} & {{z}_{2}} \\\ {{x}_{3}} & {{y}_{3}} & {{z}_{3}} \\\ \end{matrix} \right|$ is given along column 1 as
$={{x}_{1}}\left( {{y}_{2}}{{z}_{3}}-{{z}_{3}}{{y}_{2}} \right)-{{x}_{2}}\left( {{y}_{1}}{{z}_{3}}-{{y}_{3}}{{z}_{1}} \right)+{{x}_{3}}\left( {{y}_{1}}{{z}_{2}}-{{y}_{2}}{{z}_{1}} \right)..........................\left( ii \right)$
Hence, we can calculate the expansion of the given determinant in equation (i) along column 1 as
${{x}^{2}}\left( 2\times 4-1\times 1 \right)-0\left( x\times 4-1\times 1 \right)+3\left( x\times 1-2\times 1 \right)=28$
On simplifying the terms of the above equation as
$\begin{aligned} & {{x}^{2}}\left( 8-1 \right)-0+3\left( x-2 \right)=28 \\\ & 7{{x}^{2}}+3x-6=28 \\\ & 7{{x}^{2}}+3x-34=0.............................\left( iii \right) \\\ \end{aligned}$
As we know the roots of the quadratic $A{{x}^{2}}+Bx+C=0$ is given by the quadratic formula as
$x=\dfrac{-B\pm \sqrt{{{B}^{2}}-4AC}}{2A}...............................\left( iv \right)$
Now, we can compare the quadratic $7{{x}^{2}}+3x-34=0$ with the general expression of quadratic i.e. $A{{x}^{2}}+Bx+C=0$
So, we get
A = 7, B = 3, C = - 34
Hence, roots of the quadratic of equation (iii) are given as
$\begin{aligned} & x=\dfrac{-3\pm \sqrt{{{\left( 3 \right)}^{2}}-4\times 7\times -34}}{14} \\\ & x=\dfrac{-3\pm \sqrt{9+952}}{14} \\\ & x=\dfrac{-3\pm \sqrt{961}}{14} \\\ \end{aligned}$
We know the value of $\sqrt{961}$ is given as 31. So, we get value of x as
$x=\dfrac{-3\pm 31}{14}$
So, we get the values of x by taking ‘+’ and ‘-‘ sign individually as
$\begin{aligned} & x=\dfrac{-3+31}{14},\dfrac{-3-31}{14} \\\ & x=\dfrac{28}{14},\dfrac{-34}{14} \\\ & x=2,\dfrac{-17}{7} \\\ \end{aligned}$
Hence, the integral value of x is given as x = 2.
So, the answer is 2.

Note: Another method to get the values of x from the quadratic $7{{x}^{2}}+3x-34=0$ would be factorization method, where we have to apply mid-term splitting, so that their product is equal to the product of the first and last term of the quadratic. So, we can write
$\begin{aligned} & 7{{x}^{2}}+3x-34=0 \\\ & 7{{x}^{2}}+17x-14x-34=0 \\\ & x\left( 7x+17 \right)-2\left( 7x+17 \right)=0 \\\ & \left( x-2 \right)\left( 7x+17 \right)=0 \\\ & x=2,\dfrac{-17}{7} \\\ \end{aligned}$
So, it can be another approach. One may go wrong with the values of A, B and C while comparing the quadratic $7{{x}^{2}}+3x-34=0$ with $A{{x}^{2}}+Bx+C=0$. So, take care with this step. Quadratic formula for $A{{x}^{2}}+Bx+C=0$ is given as
$x=\dfrac{-B\pm \sqrt{{{B}^{2}}-4AC}}{2A}$
And get the value of ‘x’ which is an integer.