JEE MAIN - Mathematics (2022 - 25th June Evening Shift - No. 2)
Let a, b $$\in$$ R be such that the equation $$a{x^2} - 2bx + 15 = 0$$ 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 :
37
58
68
92
Explanation
$$a{x^2} - 2bx + 15 = 0$$ has repeated root so $${b^2} = 15a$$ and $$\alpha = {{15} \over b}$$
$$\because$$ $$\alpha$$ is a root of $${x^2} - 2bx + 21 = 0$$
So $${{225} \over {{b^2}}} = 9 \Rightarrow {b^2} = 25$$
Now $${\alpha ^2} + {\beta ^2} = {(\alpha + \beta )^2} - 2\alpha \beta = 4{b^2} - 42 = 100 - 42 = 58$$
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