JEE Main 2022 · 25 Jun · Shift 2 · Mathematics Q2

← JEE Main 2022 · 25 Jun · Shift 2

MathematicsSingle correct+4 / -1Question 2 of 90

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

Answer and worked solution

Official answer: B

Repeated root. $4b^2=60a$ , so $a=\frac{b^2}{15}$ and $\alpha=\frac ba=\frac{15}b$ .

$\alpha$ is a root of $x^2-2bx+21=0$ : $\frac{225}{b^2}-30+21=0$ , so $b^2=25$ ; take $b=5$ , $\alpha=3$ .

Other root. $\beta=2b-\alpha=7$ , so $\alpha^2+\beta^2=9+49=58$ (the same for $b=-5$ ).

(B)

Topics:Roots and coefficients Nature of roots

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JEE Main 2022 · 25 Jun · Shift 2, Mathematics question 2