Question

If – 4 is a root of the quadratic equation ${{x}^{2}}+px-4=0$ and the quadratic equation ${{x}^{2}}+px+k=0$ has equal roots, find the value of p and k.

Answer 1

Hint:First of all, find the value of p by substituting x = – 4 in equation (i) that is ${{x}^{2}}+px-4=0$. Now substitute the value of p in equation (ii), that is ${{x}^{2}}+px+k=0$ , and use ${{D}^{2}}={{b}^{2}}-4ac=0$ to get the value of k.

Complete step-by-step answer:
Here we are given that – 4 is a root of the quadratic equation ${{x}^{2}}+px-4=0$ and the quadratic equation ${{x}^{2}}+px+k=0$ has equal roots. We have to find the value of p and k. Before proceeding with the question, let us talk about the zeroes of the quadratic equation.
Zeroes: Zeroes or roots of quadratic equations are the value of variable say x in the quadratic equation $a{{x}^{2}}+bx+c$ at which the equation becomes zero.
We can also predict the nature of the roots of the quadratic equation using its coefficients as follows.
For the quadratic equation, $a{{x}^{2}}+bx+c=0$. If the roots are real and distinct then $d={{b}^{2}}-4ac>0$. If the roots are real and equal, then $d={{b}^{2}}-4ac=0$ and if roots are imaginary then $d={{b}^{2}}-4ac<0$.
Now, let us consider the question. Here, we are given that – 4 is the roots of the quadratic equation ${{x}^{2}}+px-4$. So, x = – 4 will satisfy this equation. So by substituting x = – 4 in this equation, we get,
${{\left( -4 \right)}^{2}}+p\left( -4 \right)-4=0$
$16-4p-4=0$
$12-4p=0$
$-4p=-12$
$4p=12$
So, we get, $p=\dfrac{12}{4}=3....\left( i \right)$
Now we are given that the quadratic equation ${{x}^{2}}+px+k=0$ has equal roots. We know that when a quadratic equation has equal roots, then the value of its discriminant $(d)={{b}^{2}}-4ac=0$. So by comparing the equation ${{x}^{2}}+px+k=0$ by standard quadratic equation $a{{x}^{2}}+bx+c$, we get, a = 1, b = p and c = k.
So, we get $d={{b}^{2}}-4ac={{p}^{2}}-4\left( 1 \right)\left( k \right)$
We know that here, d = 0. So, we get, ${{p}^{2}}-4k=0$. By substituting the value of p from equation (i), we get,
${{\left( 3 \right)}^{2}}-4k=0$
$9-4k=0$
$-4k=-9$
$k=\dfrac{-9}{-4}$
$\Rightarrow k=\dfrac{9}{4}$
Hence, we get the values of p and k as 3 and $\dfrac{9}{4}$ respectively.

Note: In this question, students must remember the value of the discriminant that is ${{b}^{2}}-4ac$ and its value according to the roots of the quadratic equation. Also, students can verify their answer by substituting the value of p and k in the given equations checking if the said conditions are true or not.