Question

Find the value of $p$ for which the quadratic equation $(p+1){{x}^{2}}-6(p+1)x+3(p+q)=0,p\ne -1$ has equal roots. Hence, find the roots of the equation.

Answer 1

Hint: We need to find the value of $p$ and also the roots. It is given that the roots are equal. So the discriminant, $D={{b}^{2}}-4ac=0$ . From the given quadratic equation, substitute for $a,b,c$ . Now by solving the value of $p$ is obtained. Substitute this value in the given quadratic equation to get a simplified quadratic equation. Factorize this to get the roots.

Complete step by step answer:
We need to find value of $p$ for which the quadratic equation $(p+1){{x}^{2}}-6(p+1)x+3(p+q)=0...(a)$ and also the roots.
It is given that the quadratic equation has equal roots.
So let us consider the discriminant, $D={{b}^{2}}-4ac$ .
When the given polynomial is compared to the standard form $a{{x}^{2}}+bx+c=0$ , we get
$a=(p+1),b=-6(p+1),c=3(p+q)$ .
Given that the quadratic equation has equal roots.
Therefore, discriminant, $D={{b}^{2}}-4ac=0$
$\Rightarrow {{b}^{2}}=4ac$
Substituting the values of $a,b,c$ in the above equation, we get
${{\left( -6(p+1) \right)}^{2}}=4(p+1)3(p+q)$
Taking the squares of the LHS gives
$36{{(p+1)}^{2}}=4(p+1)3(p+q)$
$(p+1)$ can be cancelled from both sides. Thus we get
$36(p+1)=12(p+q)$
Now expand both sides.
$36p+36=12p+12q$
Collecting the constants together, we get
$24p-12q=-36$
Taking $12$ common from LHS and the cancelling from RHS, we get
$2p-q=-3$
From this, we can get the value of $q$ as
$q=2p+3...(i)$
Or $p=\dfrac{q-3}{2}$ .
Now substitute $(i)$ in equation $(a)$ .
$(p+1){{x}^{2}}-6(p+1)x+3(p+2p+3)=0$
Simplifying, we get $(p+1){{x}^{2}}-6(p+1)x+3(3p+3)=0$
$\Rightarrow (p+1){{x}^{2}}-6(p+1)x+9(p+1)=0$
Taking $(p+1)$ outside, we get
${{x}^{2}}-6x+9=0$
Now let us find the roots of this equation. Using splitting method, we get
${{x}^{2}}-3x-3x+9=0$
Taking the common terms outside, we will get
$x(x-3)-3(x-3)=0$
Again, taking the common terms, we get
$(x-3)(x-3)=0$
This means that $(x-3)=0$ and $(x-3)=0$ .
Therefore, $x=3,3$ .

Note:
 The value of $p$ will be in terms of $q$. You can substitute either the value of $p$ or $q$ in the given quadratic equation. Either way, the simplified equation will be the same. The factors of the simplified equation can also be obtained using $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ . Note that the roots are equal as specified in the question.