Question

If α and β are the roots of the equation ax² + bx + c =0, express the roots of the equation a³x² - ab²x + b²c = 0 in terms of α and β.

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Answer 1

Given,

α and β are the roots of the equation ax²+bx+c=0.

To find,

the roots of the equation a³x² - ab²x + b²c = 0(in terms of α and β).

Solution,

We know that,

$\red\mapsto{\bf{\green{Sum\:of\:zeroes=\alpha + \beta = \frac{ - b}{a} }}}$

$\green\mapsto{\bf{\red{Product\:of\:zeroes=\:\alpha \beta = \frac{c}{a}}}}$

By dividing the equation by a³, we'll get:-

=> $\bf{x^2+ \frac{b}{a} \times \frac{c}{a}\times\: x + (\frac{c}{a})^3= 0}$

=> x² - (α+ β)(αβ)x + (αβ)³= 0

=> x²- α²βx - αβ²x + (αβ)³ = 0

=> x(x - α²β) - αβ²(x + α²β)=0

=> (x-α²β) (x-αβ²) = 0

α²β and αβ² are the roots of the equation a³x² - ab²x + b²c = 0 in terms of α and β.

Answer 2

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α and β are the roots of the equation.

⇒ ax² + bx + c = 0.

As we know that,

Sum of the zeroes of the quadratic polynomial.

⇒ α + β = - b/a.

Products of the zeroes of the quadratic polynomial.

⇒ αβ = c/a.

Roots of the equation.

⇒ a³x² - ab²x + b²c = 0.

As we know that,

Let, γ and δ are the roots of the equation.

⇒ a³x² - ab²x + b²c = 0.

Sum of the zeroes of the quadratic polynomial.

⇒ γ + δ = - b/a.

⇒ γ + δ = - (-ab²)/a³ = ab²/a³.

⇒ γ + δ = (b²/a²) = (-b/a)² = (α + β)².

Products of the zeroes of the quadratic polynomial.

⇒ γδ = c/a.

⇒ γδ = (b²c/a³) = (b²/a²)(c/a).

⇒ γδ = (b/a)²(c/a) = (α + β)².(αβ).

⇒ (γ - δ)² = (γ + δ)² - 4γδ.

⇒ (γ - δ)² = [(α + β)²]² - 4(αβ)(α + β)².

⇒ (γ - δ)² = (α + β)⁴ - 4(αβ)(α + β)².

⇒ (γ - δ) = (α² - β²). - - - - - (1).

⇒ (γ + δ) = (α + β)².

⇒ (γ + δ) = α² + β² + 2αβ. - - - - - (2).

Adding equation (1) and (2), we get.

⇒ 2γ = 2α² + 2αβ.

⇒ γ = α² + αβ.

Put the values of γ = α² + αβ in equation (1), we get.

⇒ (γ - δ) = (α² - β²). - - - - - (1).

⇒ (α² + αβ) - δ = (α² - β²)

⇒ (α² + αβ) - (α² - β²) = δ.

⇒ α² + αβ - α² + β² = δ.

⇒ β² + αβ = δ.

Values of γ = α² + αβ. and β² + αβ = δ.

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hope it helps you:)

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