QJEEM22Maaf26: JEE Main 2022 Mathematics - Class 11 JEEM22Maaf26 Solution: Let a, b ∈ R be such that the equation a x 2 − 2 b x + 15 = 0…

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QJEEM22Maaf26: JEE Main 2022 Mathematics - Class 11 JEEM22Maaf26 Solution: Let a, b ∈ R be such that the equation a x 2 − 2 b x + 15 = 0…

Let a, b $\in$ R be such that the equation $a x^{2} - 2 b x + 15 = 0$ has a repeated root $α$ . If $α$ and $β$ are the roots of the equation $x^{2} - 2 b x + 21 = 0$ , then $α^{2} + β^{2}$ is equal to :

Multiple Choice Options

A 37
B 58
C 68
D 92

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Accepted Answer

$a x^{2} - 2 b x + 15 = 0$ has repeated root so $b^{2} = 15 a$ and $α = \frac{15}{b}$

$∵$ $α$ is a root of $x^{2} - 2 b x + 21 = 0$

So $\frac{225}{b^{2}} = 9 \Rightarrow b^{2} = 25$

Now $α^{2} + β^{2} = (α + β)^{2} - 2 α β = 4 b^{2} - 42 = 100 - 42 = 58$

QJEEM22Maaf26: JEE Main 2022 Mathematics - Class 11 JEEM22Maaf26 Solution: Let a, b ∈ R be such that the equation a x 2 − 2 b x + 15 = 0…

Multiple Choice Options

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