Question

If one root of the equation ${x^2} + px + 12 = 0$ is 4 while the equation ${x^2} + px + q = 0$ has equal roots, the value of $q$ is
A ) $\dfrac{{49}}{4}$
B ) $\dfrac{4}{{49}}$
C ) $4$
D ) None of these

Answer 1

Hint: Use the concept of factor theorem for one equation and nature of roots related to the value of the discriminant in case of equal roots in case of the other quadratic equation to find out the values of the variables p and q.

Complete step by step solution:
We have given that one root of the equation ${x^2} + px + 12 = 0$ is 4 and the equation ${x^2} + px + q = 0$ has equal roots.
We know that if $\alpha $ is a root of any polynomial $f\left( x \right)$ , then from factor theorem we get $f\left( \alpha \right) = 0$
Thus, if 4 is a root of the equation ${x^2} + px + 12 = 0$, we get
${\left( 4 \right)^2} + p\left( 4 \right) + 12 = 0$
Solving the above equation, we get
$\begin{array}{l}
16 + 4p + 12 = 0\\
 \Rightarrow 4p = - 28\\
 \Rightarrow p = - 7.......\left( 1 \right)
\end{array}$
We know from the nature of a quadratic equation, that if the quadratic equation${x^2} + px + q = 0$ has equal roots, it implies that the discriminant of the quadratic equation is equal to zero.
Hence, we get , $D = 0 \Rightarrow {p^2} - 4q = 0$
Using the value of p from the equation (1) we get,
${\left( { - 7} \right)^2} - 4q = 0$
Solving further, we get the value
$q = \dfrac{{49}}{4}$

Hence, the first option is the correct answer.

Note: The Factor Theorem states that (x – a) is a factor of the polynomial f(x) if and only if f(a) = 0. For any quadratic equation, the value of its discriminant gives the nature of its roots. If the discriminant is greater than zero, then the roots are real and distinct. If the discriminant is equal to zero, then the roots are real and equal. If the discriminant is less than zero, then the roots are not real (they are complex).