JEE Main 2022 · 27 Jun · Shift 2 · Mathematics Q22

← JEE Main 2022 · 27 Jun · Shift 2

MathematicsInteger+4 / -1Question 22 of 90

Let $\alpha,\beta$ be the roots of the equation $x^2-4\lambda x+5=0$ and $\alpha,\gamma$ be the roots of the equation $x^2-\left(3\sqrt2+2\sqrt3\right)x+7+3\lambda\sqrt3=0$ , $\lambda>0$ . If $\beta+\gamma=3\sqrt2$ , then $(\alpha+2\beta+\gamma)^2$ is equal to ______.

Your answer

Answer and worked solution

Official answer: 98

Sums. $\alpha+\beta=4\lambda$ and $\alpha+\gamma=3\sqrt2+2\sqrt3$ ; with $\beta+\gamma=3\sqrt2$ , $\alpha=2\lambda+\sqrt3$ and $\beta=2\lambda-\sqrt3$ .

Product. $\alpha\beta=4\lambda^2-3=5$ , so $\lambda=\sqrt2$ : $\alpha=2\sqrt2+\sqrt3$ , $\beta=2\sqrt2-\sqrt3$ , $\gamma=\sqrt2+\sqrt3$ .

Value. $\alpha+2\beta+\gamma=7\sqrt2$ , squared 98.

Answer: 98

Topics:Roots and coefficients

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JEE Main 2022 · 27 Jun · Shift 2, Mathematics question 22