Question

if the sum of the roots of the quadratic equation ax^2+bx+c=0 is equal to the sum of the square of their reciprocal then prove that 2a^2c=c^2b+b^2a​

Answer 1

Answer:

I hope it will help you

Step-by-step explanation:

Let X and Y be the roots of the given eqn

so,

X+Y= -(b/a)

X*Y= c/a

given: X+Y=(1/X)² + (1/Y)²

X+Y=(X²+Y²)/(X²*Y²)

(X+Y)²=X²+Y²+2X*Y

X²+Y²=(X²+Y²)-2X*Y

-b/a={-(b/a)² - 2(c/a)}/(c²/a²)

-b/a={(b²-2ac)/a²}/(c²/a²)

(-bc²)/a=b²-2ac

-c²b=b²a-2a²c

[2a²c=c²b+b²a] HENCE PROVED!!!