If α,β are the roots of ax2 + 2bx + c = 0 and α + δ, β+ δ are the roots of Ax2 + 2Bx + C = 0 , then b2−acB2−AC=

here, α+β=−ba,αβ=ca and (α+δ)+(β+δ)=−2BA,(α+δ)(β+δ)=CA Now,α−β=(α+δ)−(β+δ)(α−β)2={(α+δ)−(β+δ)}2(α+β)2−4αβ={(α+δ)+(β+δ)}2−4(α+δ)(β+δ)4b2a2−4ca=4B2A2−4CAb2−aca2=B2−ACA2b2−acB2−AC=aA2Hence, the result.

If α,β are the roots of ax2 + 2bx + c = 0 and α + δ, β+ δ are the roots of Ax2 + 2Bx + C = 0 , then b2−acB2−AC=

here, α+β=−ba,αβ=ca and (α+δ)+(β+δ)=−2BA,(α+δ)(β+δ)=CA Now,α−β=(α+δ)−(β+δ)(α−β)2={(α+δ)−(β+δ)}2(α+β)2−4αβ={(α+δ)+(β+δ)}2−4(α+δ)(β+δ)4b2a2−4ca=4B2A2−4CAb2−aca2=B2−ACA2b2−acB2−AC=aA2Hence, the result.

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If $α, β$ are the roots of ax² + 2bx + c = 0 and $α$ + $δ$ , $β$ + $δ$ are the roots of Ax² + 2Bx + C = 0 , then $\frac{b^{2} - a c}{B^{2} - A C} =$

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$\frac{b^{2} - a c}{B^{2} - A C} = (\frac{a}{A})$

$\frac{b^{2} + a c}{B^{2} - A C} = (\frac{a}{A})$

$\frac{b^{2} - a c}{B^{2} - A C} = {(\frac{a}{A})}^{2}$

$\frac{b^{2} + a c}{B^{2} + A C} = (\frac{a}{A})$

answer is C.

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Detailed Solution

here, $α + β = - \frac{b}{a}, α β = \frac{c}{a}$ and $(α + δ) + (β + δ) = - \frac{2 B}{A}, (α + δ) (β + δ) = \frac{C}{A}$ Now, $α - β = (α + δ) - (β + δ)$

$\begin{array}{l} (α - β)^{2} = {(α + δ) - (β + δ)}^{2} \\ (α + β)^{2} - 4 α β = {(α + δ) + (β + δ)}^{2} - 4 (α + δ) (β + δ) \\ \frac{4 b^{2}}{a^{2}} - \frac{4 c}{a} = \frac{4 B^{2}}{A^{2}} - \frac{4 C}{A} \\ \frac{b^{2} - a c}{a^{2}} = \frac{B^{2} - A C}{A^{2}} \end{array}$