Roots of a Quadratic equation

**Ilsa** · April 2nd 2011, 03:21 AM

Given that $3p(x^2) - 7qx + 3p = 0$ has equal roots
and $p$ and $q$ are positive,
find the ratio $p : q$
and solve the equation.

(I was able to deduce the relation $49q^2 = 36p$ , but how is the ratio further on calculated?
How are the value of $p$ and $q$ deduced in such a way that substituting their values would solve the equation? )

(The answers provided are $7 : 6$ and $1$ .)

**veileen** · April 2nd 2011, 03:34 AM

$\Delta = (-7q)^2-4\cdot 3p\cdot 3p=49q^2-36p^2$
Because the equation has equal roots $\Delta=0\Rightarrow 49q^2=36p^2\Rightarrow \frac{49}{36}=\frac{p^2}{q^2} \Rightarrow \frac{7}{6}=\frac{p}{q}$

$3px^2-7qx+3p=0\Leftrightarrow 3\frac{p}{q}x^2-7x+3\frac{p}{q}=0$ Solve it.

**DrSteve** · April 2nd 2011, 03:35 AM

Originally Posted by Ilsa

Given that $3p(x^2) - 7qx + 3p = 0$ has equal roots
and $p$ and $q$ are positive,
find the ratio $p : q$
and solve the equation.

(I was able to deduce the relation $49q^2 = 36p$ , but how is the ratio further on calculated?
How are the value of $p$ and $q$ deduced in such a way that substituting their values would solve the equation? )

(The answers provided are $7 : 6$ and $1$ .)

You have a small error. It should be $49q^2 = 36p^2$ from which it follows that $\frac{p^2}{q^2} = \frac{49}{36}$ . Can you take it from here?

**HallsofIvy** · April 2nd 2011, 04:14 AM

Another way to do this: if we call the two equal roots "r" then we must have
$3p(x- r)^2= 3p(x^2- 2rx+ r^2)= 3px^2- 6prx+ 3pr^2= 3px^2- 7qx+ 3p$
and so must have $6pr= 7q$ and $3pr^2= 3p$ . That second equation reduces to $r^2= 1$ so that r= 1 or r= -1. In order that p and q both be positive, we must have r= 1 and then we have $6p= 7q$ so that $\frac{p}{q}= \frac{7}{6}$ .

**Soroban** · April 2nd 2011, 05:05 AM

Hello, Ilsa!

$\text{The answers provided are: }7\!:\!6 \text{ and } 1.$
. . . . . . . . . . . . . . . . . . . . . . . . . . . . $\uparrow$
. . . . . . . . . . . . . . . . . . . . . . . . . . . . $?$