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.