Question

Find the value 'p' so that the equation $4x^2-8px+9=0$ has roots whose difference is 4.

Answer 1

We know that if $m$ and $n$ are the roots of a quadratic equation $a{x}^2+bx+c=0$, the sum of the roots is $m+n=-\frac{b}{a}$ and the product of the roots is $mn=\frac{c}{a}$.

Let $m$ and $n$ be the roots of the given quadratic equation $4x^2-8px+9=0$. It is given that the difference of the roots is $4$, therefore,

$m-n=4........\left(1\right)$

The equation $4x^2-8px+9=0$is in the form $a{x}^2+bx+c=0$ where $a=4,b=-8q$ and $c=9$.

The sum of the roots is:

$m+n=-\frac{b}{a}=-\frac{(-8q)}{4}=2q....\left(2\right)$

The product of the roots is $\frac{c}{a}$ that is:

$mn=\frac{c}{a}=\frac{9}{4}......\left(3\right)$

Now, we know the identity $(m+n{)}^2=(m-n{)}^2+4mn$, therefore, using equations 1,2 and 3, we have

$(m+n{)}^2=(m-n{)}^2+4mn\Rightarrow (2p{)}^2=4^2+(4\times \frac{9}{4})\Rightarrow 4p^2=16+9\Rightarrow 4p^2=25\Rightarrow p^2=\frac{25}{4}\Rightarrow p=\pm \sqrt{\frac{25}{4}}\Rightarrow p=\pm \frac{5}{2}$

Hence, the value of $p=\pm \frac{5}{2}$.