Let α , β be the roots of the equation x 2 - 4 λ x + 5 = 0 and α , γ be the roots…

Let α,β be the roots of the equation x2-4λx+5=0 and α,γ be the roots of the equation x2-32+23x+7+3λ3=0. If β+γ=32, then α+2β+γ2 is equal to

Solution

Given $\alpha$ and $\beta$ are roots of $x^{2}-4\lambda x+5=0$ So, $\alpha+\beta=4\lambda$ and $\alpha\beta=5$ .......(i) And $\alpha$ and $\gamma$ are roots of $x^{2}-\left(3\sqrt{2}+2\sqrt{3}\right)x+\left(7+3\lambda\sqrt{3}\right)=0$ So, $\alpha+\gamma=3\sqrt{2}+2\sqrt{3}$ and $\alpha\gamma=7+3\lambda\sqrt{3}$ .........(ii) Given, if $\beta+\gamma=3\sqrt{2}$ So, from equation (i) and (ii) we get, $\alpha=2\lambda+\sqrt{3}$ and $\beta=2\lambda-\sqrt{3}$, Now by product of roots we get, $4\lambda^{2}-3=5 \Rightarrow \lambda=\sqrt{2}$ $\therefore \left(\alpha+2\beta+\lambda\right)^{2}=\left(\alpha+\beta+\beta+\lambda\right)^{2}=\left(4\lambda+3\sqrt{2}\right)^{2}=\left(4\sqrt{2}+3\sqrt{2}\right)^{2}=\left(7\sqrt{2}\right)^{2}=98$

Asked in: JEE Main 2022 (27 Jun Shift 2)

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