# If the roots of ax^2 + bx +c = 0 are in the ratio 3:4, prove that 12b^2 = 49ac?

Jan 10, 2017

$12 {b}^{2} = 49 a c$

#### Explanation:

$y = a {x}^{2} + b x + c = 0$
Reminder of the improved quadratic formula (Socratic Search)
Determinant --> $D = {d}^{2} = {b}^{2} - 4 a c$, with $d = \pm \sqrt{D}$
The 2 real roots are:
$x = - \frac{b}{2 a} \pm \frac{d}{2 a}$
$x 1 = \frac{- b + d}{2 a}$
$x 2 = \frac{- b - d}{2 a} = \frac{- \left(b + d\right)}{2 a}$
The ration $\frac{x 1}{x 2} = \frac{3}{4} = \frac{d - b}{-} \left(b + d\right)$
Cross multiply -->
- 3b - 3d = 4d - 4b
4b - 3b = 4d + 3d
b = 7d.
Square both sides -->
${b}^{2} = 49 {d}^{2} = 49 \left({b}^{2} - 4 a c\right) = 49 {b}^{2} - 196 a c$
$48 {b}^{2} = 196 a c$. Simplify by 4.
$12 {b}^{2} = 49 a c$

Jan 10, 2017

Let $\alpha \mathmr{and} \beta$ are two roots of the given quadratic equation $a {x}^{2} + b x + c = 0$

Hence $\alpha + \beta = - \frac{b}{a} \mathmr{and} \alpha \beta = \frac{c}{a}$

Again it is also given that $\frac{\alpha}{\beta} = \frac{3}{4}$
Let $\alpha = 3 k \mathmr{and} \beta = 4 k$

So $7 k = - \frac{b}{a} \mathmr{and} 12 {k}^{2} = \frac{c}{a}$

Hence $\frac{49 {k}^{2}}{12 {k}^{2}} = {\left(- \frac{b}{a}\right)}^{2} / {\left(\frac{c}{a}\right)}^{2} = {b}^{2} / \left(a c\right)$

$\implies 12 {b}^{2} = 49 a c$

Proved