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

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

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Answer 1 · 2017-01-10T05:09:04

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

Explanation:

$y = a x^{2} + b x + c = 0$ $y = ax^2 + bx + c = 0$
Reminder of the improved quadratic formula (Socratic Search)
Determinant --> $D = d^{2} = b^{2} - 4 a c$ $D = d^2 = b^2 - 4ac$ , with $d = \pm \sqrt{D}$ $d = +- sqrtD$
The 2 real roots are:
$x = - \frac{b}{2 a} \pm \frac{d}{2 a}$ $x = -b/(2a) +- d/(2a)$
$x 1 = \frac{- b + d}{2 a}$ $x1 = (-b + d)/(2a)$
$x 2 = \frac{- b - d}{2 a} = \frac{- (b + d)}{2 a}$ $x2 = (-b - d)/(2a) = (-(b + d))/(2a)$
The ration $\frac{x 1}{x 2} = \frac{3}{4} = \frac{d - b}{- (b + d)}$ $(x1)/(x2) = 3/4 = (d - b)/-(b + d)$
Cross multiply -->
- 3b - 3d = 4d - 4b
4b - 3b = 4d + 3d
b = 7d.
Square both sides -->
$b^{2} = 49 d^{2} = 49 (b^{2} - 4 a c) = 49 b^{2} - 196 a c$ $b^2 = 49d^2 = 49(b^2 - 4ac) = 49b^2 - 196 ac$
$48 b^{2} = 196 a c$ $48b^2 = 196ac$ . Simplify by 4.
$12 b^{2} = 49 a c$ $12b^2 = 49ac$

Answer 2 · 2017-01-10T11:12:03

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

Hence $α + β = - \frac{b}{a} and α β = \frac{c}{a}$ $alpha+beta=-b/a and alphabeta=c/a$

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

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

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

$\Rightarrow 12 b^{2} = 49 a c$ $=>12b^2=49ac$

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

2 Answers

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