Question

Find the value of P so that the quadratic equation ${\text{px(x - 3) + 9 = 0}}$ has two equal roots.

Answer 1

Hint: We know that the quadratic equation $a{x^2} + bx + c = 0$ has two equal roots only when discriminant D=0 i.e. ${b^2} - 4ac = 0$
We use this fact to find the value of P.

Complete step-by-step answer:
Given, the equation ${\text{px(x - 3) + 9 = 0 }} \Rightarrow {\text{p}}{{\text{x}}^2} - 3px + 9 = 0$
We know that the quadratic equation $a{x^2} + bx + c = 0$ has two equal roots only when discriminant D=0 i.e. ${b^2} - 4ac = 0$
Here discriminant D=${( - 3p)^2} - 4.p.9 = 9{p^2} - 36p$
∴ the equation ${\text{p}}{{\text{x}}^2} - 3px + 9 = 0$ will have two equal roots if only D=0
i.e. $9{p^2} - 36p = 0$
On taking 9p common we get,
$ \Rightarrow 9p(p - 4) = 0$
So, we have either $(p - 4) = 0$or $9p = 0$,
Therefore either ${\text{p = 0 or p = 4}}$
Therefore, the equation ${\text{px(x - 3) + 9 = 0}}$ will have two equal roots p=0 or p=4.

Note: Let us consider a quadratic function f(x) = $a{x^2} + bx + c$
Therefore, the roots of the function are given by $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
We call the term ${b^2} - 4ac$ the discriminant. The discriminant tells us about the nature of the roots.
1.If discriminant D>0, the roots are real.
2.If discriminant D=0, the roots are real and equal.
3.If discriminant D<0, the roots are imaginary, therefore no real root exists.