If - and - are the roots of a quadratic equation satisfying the conditons -4 and -1-1- a 2-7- a 2-4- - - a - R- For what values of - a -will the quadratic equation have equal roots- A- -3B- -5- -3D- -5

The correct options are A √(3) C √(3)α (β 1)+β(α 1)(α 1)(β 1)=a^2 7a^2 4 ⇒2αβ (α+β)αβ (α+β)+1=a^2 7a^2 4 ⇒8 (α+β)5 (α+β)=a^2 7a^2 4 [∵αβ =4] 1+35 (α +β ) ...

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Byju's Answer

Standard XIII

Mathematics

Relation between Roots and Coefficients for Quadratic

If α and β ar...

Question

If $α$ and $β$ are the roots of a quadratic equation satisfying the conditons $α β = 4$ and $\frac{α}{α - 1} + \frac{β}{β - 1} = \frac{a^{2} - 7}{a^{2} - 4}, α, β, a \in R$ . For what values of $^{'} a^{'}$ will the quadratic equation have equal roots?

\sqrt{3}

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\sqrt{5}

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- \sqrt{3}

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- \sqrt{5}

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Solution

The correct options are
A $\sqrt{3}$
C $- \sqrt{3}$
$\frac{α (β - 1) + β (α - 1)}{(α - 1) (β - 1)} = \frac{a^{2} - 7}{a^{2} - 4}$

$\Rightarrow \frac{2 α β - (α + β)}{α β - (α + β) + 1} = \frac{a^{2} - 7}{a^{2} - 4}$

$\Rightarrow \frac{8 - (α + β)}{5 - (α + β)} = \frac{a^{2} - 7}{a^{2} - 4}$
$[∵ α β = 4]$

$1 + \frac{3}{5 - (α + β)} = 1 - \frac{3}{a^{2} - 4} \Rightarrow (α + β) = a^{2} + 1 \dots (1)$
For roots to be equal
$α β = 4 \Rightarrow α^{2} = 4 \Rightarrow α = \pm 2$
Using equation $(1)$ ,
$2 α = a^{2} + 1 \Rightarrow a = \pm \sqrt{2 α - 1} \Rightarrow a = \pm \sqrt{3}$

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If α and β are the roots of a quadratic equation satisfying the conditons αβ=4 and αα−1+ββ−1=a2−7a2−4,α,β,a∈R. For what values of ′a′ will the quadratic equation have equal roots?

If $α$ and $β$ are the roots of a quadratic equation satisfying the conditons $α β = 4$ and $\frac{α}{α - 1} + \frac{β}{β - 1} = \frac{a^{2} - 7}{a^{2} - 4}, α, β, a \in R$ . For what values of $^{'} a^{'}$ will the quadratic equation have equal roots?