Question

1) if the roots of the equation ax^2+bx+c=0 are in the ratio 2:3, then find the condition.​

Answer 1

Answer:

The roots of the equation ax^2 + bx + c = 0 are in the ratio of 2 :3, then what is the relation between a, b and c?

The roots of the equation ax^2 + bx + c = 0 are

x1 =[-b +(b^2 -4ac)^0.5]/2a, and

x2 = [-b -(b^2 -4ac)^0.5]/2a.

Now it says x2:x1::2:3, or

{[-b -(b^2 -4ac)^0.5]/2a}/{[-b +(b^2 -4ac)^0.5]/2a} = 2/3, or

3{[-b -(b^2 -4ac)^0.5]/2a} = 2{[-b +(b^2 -4ac)^0.5]/2a}, or

-3b -3(b^2 -4ac)^0.5 = -2b +2(b^2 -4ac)^0.5, or

-3b +2b = 2(b^2 -4ac)^0.5 + 3(b^2 -4ac)^0.5, or

-b = 5(b^2 -4ac)^0.5, or

-b/5 = (b^2 -4ac)^0.5. Squaring both sides we get

b^2/25 = b^2 - 4ac, or

b^2 = 25(b^2 - 4ac), or

24b^2 = 100ac

Explanation:

Answer 2

Answer:

$\large{\bold{8b^2=25ac}}$

Step-by-step explanation:

$ax^2+bx+c=0$

$Given \quad \alpha:\beta=2:3$

$\frac{\alpha}{\beta}=\frac{2}{3}$

WKT

$\frac{\alpha}{\beta}=\frac{-b+\sqrt{b^2-4ac}}{-b-\sqrt{b^2-4ac}}$

$\frac{2}{3}=\frac{-b+\sqrt{b^2-4ac}}{-b-\sqrt{b^2-4ac}}$

$=>-2b-2\sqrt{b^2-4ac}=-3b+3\sqrt{b^2-4ac}$

$b=5\sqrt{b^2-4ac}$

Squaring on Both Sides

$b^2=25(b^2-4ac)$

$24b^2=100ac$

$8b^2=25ac$

HOPE IT HELPS