Question

The sum of the real roots of the equation is
$\left| \begin{matrix}
   x & -6 & -1 \\
   2 & -3x & x-3 \\
   -3 & 2x & x+2 \\
\end{matrix} \right|=0$
$\begin{align}
  & \text{a) 6} \\
 & \text{b) 1} \\
 & \text{c) 0} \\
 & \text{d) -4} \\
\end{align}$

Answer 1

Hint:Now we will first evaluate the determinant and form the cubic equation. As we know that for any equation in the form of $a{{x}^{3}}+b{{x}^{2}}+cx+d=0$ the sum of roots is given by $-\dfrac{b}{a}$ that is negative of coefficient of ${{x}^{2}}$ divided by the coefficient of ${{x}^{3}}$. Hence we will compare the equation with $a{{x}^{3}}+b{{x}^{2}}+cx+d=0$ and then find $-\dfrac{b}{a}$ and hence sum of roots.

Complete step by step answer:
Now we are given that
$\left| \begin{matrix}
   x & -6 & -1 \\
   2 & -3x & x-3 \\
   -3 & 2x & x+2 \\
\end{matrix} \right|=0$
Now we will evaluate the determinant with respect to first row, hence we will get
x(–3x(x + 2) – 2x(x – 3)) – (–6)(2(x + 2) –(–3)(x – 3)) – 1(2(2x) – (–3)(x – 3)) = 0
$\begin{align}
  & x(-3{{x}^{2}}-6x-2{{x}^{2}}+6x)+6\left( 2x+4+3x-9 \right)-1\left( 4x+3x-9 \right) \\
 & \Rightarrow x\left( -5{{x}^{2}} \right)+6\left( 5x-5 \right)-1\left( 7x-9 \right) \\
 & \Rightarrow -5{{x}^{3}}+30x-30-7x+9 \\
 & \Rightarrow -5{{x}^{3}}+23x-21 \\
\end{align}$
Now we know that for any equation in the form of $a{{x}^{3}}+b{{x}^{2}}+cx+d=0$ the sum of roots is given by $-\dfrac{b}{a}$
Hence comparing the equation $-5{{x}^{3}}+23x-21$ with the equation $a{{x}^{3}}+b{{x}^{2}}+cx+d=0$ we get a = - 5, b = 0, c = 12 and d = - 21.
Hence now $-\dfrac{b}{a}=-\dfrac{0}{-5}$
Now we know that when we divide any number to 0 we get 0.
Hence we have $\dfrac{0}{-5}=0$
Hence $-\dfrac{b}{a}=0$
Hence the sum off roots is 0.
Hence option c is the correct option.

Note:
 Now for any polynomial of degree n we have that the sum of roots is negative ratio of the highest power to the second highest power. Hence we get the sum of roots = $-\dfrac{\text{coefficient of }{{\text{x}}^{n}}}{\text{coeffiecient of }{{\text{x}}^{n-1}}}$ .
Once we have the equation we can also try to find the roots. First, we will substitute some values in the cubic equation to get the first root. Let us say we have b as our first root then we will divide the cubic equation with (x-b) to get a quadratic equation. Now we can find the two roots of the quadratic equation easily. And hence once we have all the roots we can add them to find sum of roots.