Question

If one root of the equation $x^2+px+q=0$ is 3 times the other prove that $3p^2=16q$.

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 $x^2+px+q=0$. It is given that one of the root is three times the other, therefore,

$m=3n........\left(1\right)$

The equation $x^2+px+q=0$is in the form $a{x}^2+bx+c=0$ where $a=1,b=p$ and $c=q$.

Using equation 1, the sum of the roots is:

$m+n=-\frac{b}{a}=-\frac{p}{1}=-p\Rightarrow m+n=-p\Rightarrow 3n+n=-p\Rightarrow 4n=-p\Rightarrow n=-\frac{p}{4}....\left(2\right)$

Using equation 1, the product of the roots is $\frac{c}{a}$ that is:

$mn=\frac{c}{a}=\frac{q}{1}=q$

$\Rightarrow n=-p\Rightarrow (3n\times n)=q\Rightarrow 3n^2=q......\left(3\right)$

Now, substitute equation 2 in equation 3 as follows:

$(3n\times n)=q\Rightarrow 3{(-\frac{p}{4})}^2=q\Rightarrow (3\times \frac{p^2}{16})=q\Rightarrow \frac{3p^2}{16}=q\Rightarrow 3p^2=16q$

Hence, the value of $3p^2=16q$.