Question

If 5 is one of the root of the equation $\left| {\begin{array}{*{20}{c}}
  x&3&7 \\
  2&x&{ - 2} \\
  7&8&x
\end{array}} \right| = 0$, then the other two roots of the equation are:
(A) -2, -7
(B) -2, 7
(C) 2, -7
(D) 2, 7

Answer 1

Hint: Expand the determinant using any row or column. This will give us a cubic equation and its one root is already given i.e. 5. So $\left( {x - 5} \right)$ will be a factor of it. Thus the cubic equation can be written as the product of $\left( {x - 5} \right)$ and a quadratic expression which is its other factor. Assume a variable quadratic expression as its other factor and solve it to get the two roots.

Complete step by step answer:
According to the question, we have been given a determinant equation and we have to find two roots of it while one root is already given.
So the determinant equation is:
$ \Rightarrow \left| {\begin{array}{*{20}{c}}
  x&3&7 \\
  2&x&{ - 2} \\
  7&8&x
\end{array}} \right| = 0$
If we expand the determinant on the left hand side using the first row, we’ll get:
$
   \Rightarrow x\left[ {{x^2} - \left( { - 16} \right)} \right] - 3\left[ {2x - \left( { - 14} \right)} \right] + 7\left[ {16 - 7x} \right] = 0 \\
   \Rightarrow {x^3} + 16x - 6x - 42 + 112 - 49x = 0 \\
   \Rightarrow {x^3} - 39x + 70 = 0{\text{ }}.....{\text{(1)}} \\
 $
Now this is a cubic polynomial on the left hand side and we have to find the roots.
One of the roots is already given in the question which is 5. So $\left( {x - 5} \right)$ will be a factor of this equation. We know that the other factor will be quadratic since it’s a cubic polynomial. And hence this cubic polynomial can be written as the product of a linear factor and a quadratic factor.
Let the quadratic factor be $\left( {a{x^2} + bx + c} \right)$, so we have:
$ \Rightarrow {x^3} - 39x + 70 = \left( {x - 5} \right)\left( {a{x^2} + bx + c} \right)$
If we expand this, we will get:
$
   \Rightarrow {x^3} - 39x + 70 = a{x^3} + b{x^2} + cx - 5a{x^2} - 5bx - 5c \\
   \Rightarrow {x^3} - 39x + 70 = a{x^3} + \left( {b - 5a} \right){x^2} + \left( {c - 5b} \right)x - 5c \\
 $
Comparing coefficients on both sides of the equation, we’ll get:
$ \Rightarrow a = 1,{\text{ }}b - 5a = 0,{\text{ }}c - 5b = - 39{\text{ and }} - 5c = 70$
Simplifying these variable, we have:
$ \Rightarrow a = 1,{\text{ }}b = 5,{\text{ and }}c = - 14$
Thus the quadratic expression $\left( {a{x^2} + bx + c} \right)$ becomes $\left( {{x^2} + 5x - 14} \right)$ and it is the other factor of our cubic expression. From equation (1), we have:
$ \Rightarrow {x^3} - 39x + 70 = \left( {x - 5} \right)\left( {{x^2} + 5x - 14} \right) = 0$
To find the other two roots of the cubic equation we have to find the roots of its quadratic factor. So we have to solve the equation:
$ \Rightarrow \left( {{x^2} + 5x - 14} \right) = 0$
We will factorize this quadratic expression further by splitting the middle terms into two terms as shown below:
$ \Rightarrow {x^2} + 7x - 2x - 14 = 0$
Taking out common terms, we have:
$
   \Rightarrow x\left( {x + 7} \right) - 2\left( {x + 7} \right) = 0 \\
   \Rightarrow \left( {x - 2} \right)\left( {x + 7} \right) = 0 \\
 $
Putting both the factors to zero, we’ll get:
$
   \Rightarrow \left( {x - 2} \right) = 0{\text{ and }}\left( {x + 7} \right) = 0 \\
   \Rightarrow x = 2{\text{ and }}x = - 7 \\
 $
Thus the other two roots of the equation $\left| {\begin{array}{*{20}{c}}
  x&3&7 \\
  2&x&{ - 2} \\
  7&8&x
\end{array}} \right| = 0$ are 2 and -7.

So, the correct answer is Option C.

Note: We can also apply the property of determinants to solve it instead of expanding it form the first step. Some of the important properties are:
(1) If all the elements of any row or any column of a determinant are zero then the value of determinant is zero.
(2) If any two rows or any two columns of a determinant are same then also the value of determinant is zero.
(3) If all the elements of any row or any column are a multiple of a rational number then that number can be taken out of determinant. For example, this is shown below:
$ \Rightarrow \left| {\begin{array}{*{20}{c}}
  {ka}&{kb}&{kc} \\
  d&e&f \\
  g&h&i
\end{array}} \right| = k\left| {\begin{array}{*{20}{c}}
  a&b&c \\
  d&e&f \\
  g&h&i
\end{array}} \right|$