Question

If \[{x^2} + ax + 10 = 0\] and \[{x^2} + bx - 10 = 0\] have a common root, then ${a^2} - {b^2}$ is equal to
A. $10$
B. $20$
C. $30$
D. $40$

Answer 1

Hint: In order to solve this type of question, first we will consider the given equations. Then, we will assume the common root of both the equations and rewrite them. Next, we will subtract both the equations and substitute the obtained value in any of the rewritten equations to get the final answer.

Formula used: If ‘a’ is a root of a quadratic equation then it will satisfy the quadratic equation.

Complete step by step solution:
We are given that,
\[{x^2} + ax + 10 = 0\]
\[{x^2} + bx - 10 = 0\]
Let the common root of both the equations be $\alpha .$
\[{\alpha ^2} + a\alpha + 10 = 0\] ………………..equation$\left( 1 \right)$
\[{\alpha ^2} + b\alpha - 10 = 0\] ………………..equation$\left( 2 \right)$
Subtract equation $\left( 2 \right)$ from $\left( 1 \right)$,
$\left( {a - b} \right)\alpha + 20 = 0$
$\alpha = \dfrac{{ - 20}}{{\left( {a - b} \right)}}$
Substituting this value in equation $\left( 1 \right)$,
\[{\left( {\dfrac{{ - 20}}{{\left( {a - b} \right)}}} \right)^2} + a\left( {\dfrac{{ - 20}}{{\left( {a - b} \right)}}} \right) + 10 = 0\]
$400 - 20a\left( {a - b} \right) + 10{\left( {a - b} \right)^2} = 0$
Simplifying it,
$40 - 2{a^2} + 2ab + {a^2} + {b^2} - 2ab = 0$
${a^2} - {b^2} = 40$
Thus, the correct option is D.

Note: The sum of roots of a quadratic equation is given by $\dfrac{{ - b}}{a}.$ The product of the roots of a quadratic equation is given by $\dfrac{c}{a}.$ If $\alpha $ is the root of ${x^2} + px + q = 0$ then it will satisfy ${\alpha ^2} + p\alpha + q = 0$ which will be followed by other roots of the quadratic equation also.