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

$\large\underline {\sf {\pink{ \pink\star {\: GIVEN: - }}}}$

$\sf{ \rightarrow \: If \: a \: and \: B are \: the \: roots \: of \: the \: equation \: } \\ \sf{ ax ^{2} + bx + c =0,}</p><p>$

$\large\underline {\sf {\red{ \pink\star {\: TO \: FIND : - }}}}$

$\sf{ \rightarrow \: express \: the \: roots \: of \: the \: equation \:} \\ \sf{ {a}^{3} {x}^{2} - ab ^{2} x + b ^{2} c = 0 \: in \: terms \: of \: } \\ \sf{a \: and \: B \: .}</p><p>$

$\large\underline {\sf {\orange{ \pink\star {\: SOLUTION \: : - }}}}$

$\sf\green{ \alpha \: and \beta \: are \: the \: roots \: of \: the \: equation. \: }$

$\sf{ \Rightarrow \: ax ^{2} + bx + c = 0.}$

$\sf \blue{As \: we \: know \: that,}$

$\sf\green{Sum \: of \: the \: zeroes \: of \: the \: quadratic } \\ \sf{ \: polynomial.} \\ </p><p> \sf{ \Rightarrow\alpha + \beta = \frac{ - b}{a} }$

$\sf\green{Products \: of \: the \: zeroes \: of \: the \: quadratic \: } \\\sf\green{ polynomial.} \\ \sf{ \Rightarrow \: \alpha \beta = \frac{c}{a} }$

$\sf\green{Roots \: of \: the \: equation}$

$\sf{ \Rightarrow \: a ^{3} \: x ^{2} ab ^{2} x + b^{2} c = 0.}$

$\sf \blue{As \: we \: know \: that,}$

$\sf\green{Let, \:y \:and\: \delta \:are \:the \:roots\: of \:the \:equation.}$

$\sf{\Rightarrow a ^{3} \: x ^{2} - ab ^{2} x + b^{2}c = 0.}$

$\sf\green{Sum \:of\: the \:zeroes \:of \:the \:quadratic\: }$

$\sf\green{polynomial.}$

$\sf{\Rightarrow \:y + \delta = \frac{-b}{a}}$

$\sf{\Rightarrow \:y + \delta=\frac{(-ab^{2})}{a^{3}} = ab^{2}/a^{3}.}$

$\sf{\Rightarrow \:y +\delta = =(\: \frac{b^{2}}{a^{2}}\: )=(\: \frac{-b}{a}^{2}\: )= (\: \alpha +\beta \: )^{2}.}$

$\sf\green{Products \:of\: the \:zeroes \:of \:the \:quadratic}$

$\sf\green{polynomial.}$

$\sf{\Rightarrow \: y \delta = \frac{c}{a}}$

$\sf{\Rightarrow \: y \delta =(\frac{b^{2}c}{a^{3}})=(\frac{b^{2}}{a^{2}}) (\frac{c}{a}) }$

$\sf{\Rightarrow \:y \delta =(\: \frac{b}{a}^{2}\: )(\frac{c}{a})= (\: \alpha +\beta \: )^{2}(\: \alpha +\beta \: ) }$

Answer 2

Answer:

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.  

Explanation:

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