Given the quadratic equation $p(x^2 +9)= -5qx$ has two equal roots, find the ratio of $p:q$. Hence, solve the quadratic equation

Question

Given the quadratic equation $p(x^2 +9)= -5qx$ has two equal roots, find the ratio of p:q. Hence, solve the quadratic equation

so this is what i got so far :

$$px^2+9p+5qx=0$$

$$(5q)^2 - 4(p)(9p)=0$$

$$25q^2 - 36p^2 =0$$

$$(5^2 q^2) - (6^2 p^2)=0$$

$$5^2 q^2 = 6^2 p^2$$

$$5q =6p$$

I might be wrong what to do next?

Answer 1

$p(x^2 + 9) + 5qx = 0 \Rightarrow px^2 + 5qx + 9p = 0$

$x= \frac{-5q \pm \sqrt{5^2q^2 -4\cdot9p^2}}{2p}$

Then you want the discriminant equal to 0. This is:

$5^2q^2 -4\cdot9p^2 = 0 \Rightarrow 5^2q^2 = 4\cdot9p^2 \Rightarrow 5q = 6p \Rightarrow p=\frac{5q}{6} \Rightarrow \frac{p}{q} = \frac{5}{6}$

Now the roots of the equation are:

$x= \frac{-5q}{2p} = \frac{-5\cdot6}{2\cdot5} = -3$

Answer 2

Answer 1 -

When we have equal roots then discriminate equals to zero. As you are solving.

Answer 2 -

We have,

5q = 6p

$\frac{p}{q} = \frac56$

Or

You can also write

$(5^2 q^2) - (6^2 p^2) = 0$

As $(5q)^2 - (6p)^2 = 0$

$(5q - 6p)(5q + 6p) = 0$

Either 5q - 6p = 0 or 5q + 6p = 0

$\frac{p}{q} = \frac56$ or $\frac{p}{q} = \frac{-5}{6}$

Answer 3 -

For equal roots we have each root equals to $\frac{-b}{2a}$

So x = $\frac{-5q}{2p}$

= $\frac{-5}{2} \times \frac{6}{5}$

x = -3