Question

If the sum of the roots of the quadratic equation ax2+bx+c=0 is equal to the sum of the square of their reciprocals, the prove that 2a2c=c2b+b2a.​

Answer 1

Step-by-step explanation:

Above answe is correct

Answer 2

$2a^2c=c^2b+b^2a$, proved.

Step-by-step explanation:

Let α and β be the given roots of quadratic equation.

The given quadratic equation:

a$x^2$ + bx + c = 0

To prove that $2a^2c=c^2b+b^2a$.

We know that,

α + β = $\dfrac{-b}{a}$ and

αβ = $\dfrac{c}{a}$

According to question,

$\alpha+\beta=\dfrac{1}{\alpha^2} +\dfrac{1}{\beta^2}$

⇒ $\alpha+\beta=\dfrac{\alpha^2+\beta^2}{\alpha^2\beta^2}$

⇒ $\alpha+\beta=\dfrac{(\alpha+\beta)^2-2\alpha\beta}{\alpha^2\beta^2}$

⇒ $\alpha+\beta=\dfrac{(\alpha+\beta)^2-2\alpha\beta}{(\alpha\beta)^2}$

Put α + β = $\dfrac{-b}{a}$ and αβ = $\dfrac{c}{a}$ , we get

$\dfrac{-b}{a}=\dfrac{(\dfrac{-b}{a})^2-2\dfrac{c}{a}}{(\dfrac{c}{a})^2}$

⇒ $\dfrac{-b}{a}=\dfrac{\dfrac{b^2}{a^2}-2\dfrac{c}{a}}{\dfrac{c^2}{a^2}}$

⇒ $-bc^2=ab^2-2a^2c$

⇒ $2a^2c=c^2b+b^2a$, proved.

Thus, $2a^2c=c^2b+b^2a$, proved.