Question

form the equation whose roots are alphasquare,betasquare where alpha, beta are roots of axsquare + bx+c =0​

Answer 1

Answer :

The required equation is   $\sf x^2a^2+(2ac-b^2)x+c^2=0$

Step-by-step explanation :

Given quadratic equation,

ax² + bx + c = 0

=> α and β are the zeroes of the given polynomial.

From the relation between zeroes and coefficients,

⇒ Sum of the zeroes = -(x coefficient)/x² coefficient

 α + β = -b/a

⇒ Product of the zeroes = constant term/x² coefficient

   αβ = c/a

________________________

We have to form a quadratic equation whose roots are α² , β²

⇒ Sum of zeroes = α² + β²

⇒ Product of zeroes = (α²)(β²) = α²β²

We know that,

(a + b)² = a² + b² + 2ab

⇒ a² + b² = (a + b)² - 2ab

Similarly,

α² + β² = (α + β)² - 2αβ

$\sf \alpha^2+\beta^2 = (\frac{-b}{a})^2-2(\frac{c}{a}) \\\\ \alpha^2+\beta^2=\frac{b^2}{a^2} -\frac{2c}{a} \\\\ \alpha^2+\beta^2=\frac{b^2-2ac}{a^2}$

The quadratic polynomial is of the form

x² - (sum of zeroes)x + (product of zeroes)

The required form of equation is

x² - (α² + β²)x + (α²β²) = 0

x² - (α² + β²)x + (αβ)² = 0

$\sf x^2-[\frac{b^2-2ac}{a^2}]x+(\frac{c}{a})^2 = 0\\\\ x^2 -\frac{(b^2-2ac)x}{a^2} + \frac{c^2}{a^2} = 0 \\\\ \frac{x^2a^2-(b^2-2ac)x+c^2 }{a^2} =0 \\\\ x^2a^2-(b^2-2ac)x+c^2=0 \times a^2 \\\\ x^2a^2-(b^2-2ac)x+c^2=0 \\\\ x^2a^2+ (2ac-b^2)x+c^2=0$

Therefore, the required quadratic equation is

$\bf x^2a^2+ (2ac-b^2)x+c^2=0$