Let a, b ∈ R be such that the equation a x ^ 2 - 2bx + 15 = 0 has...

Given $a{x}^{2} - 2bx + 15 = 0 . . .$\left(i\right)$</p><p>Has repeated roots So$D = 0$</p><p>$4{b}^{2} - 4 \times 15 \times a = 0$</p><p>$⇒ {b}^{2} = 15a . . . \left(ii\right)$</p><p>Also given </p><p><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/606b6ad2-7986-4c40-bcf4-ead8883392df-image.png" style="width: 247px; height: 65px;" /></p><p>Now$\alpha$will satisfy both quadratic</p><p>$a{x}^{2} - 2bx + 15 = 0 \text{ and } {x}^{2} - 2bx + 21 = 0$</p><p>Putting the value we get</p><p>$a{\alpha}^{2} - 2b\alpha + 15 = 0{\alpha}^{2} - 2b\alpha + 21 = 0 - + - - - - - - - - - - - \left(a - 1\right){\alpha}^{2} = 6$</p><p>${\alpha}^{2} = \dfrac{6}{a - 1}$</p><p>Now in equation (1) product of Root${\alpha}^{2} = \dfrac{15}{a}$</p><p>So$\dfrac{15}{a} = \dfrac{6}{a - 1}$$⇒ 2a = 5a - 5 ⇒ a = \dfrac{5}{3}$

Now ${b}^{2} = 15a ⇒ {b}^{2} = 15 \times \dfrac{5}{3} ⇒ {b}^{2} = 25$

So $b = \pm 5$

Now in quadratic ${x}^{2} - 2bx + 21 = 0$

Putting the value of $b$ we get

${x}^{2} - 10x + 21 = 0 ⇒ \left(x - 7\right)\left(x - 3\right) = 0$

So $x = 3 or$7$.</p><p>Or </p><p>${x}^{2} + 10x + 21 = 0 ⇒ x = - 3orx = - 7$</p><p>So$\alpha = \pm 3 \text{ and } \beta = \pm 7$</p><p>So${\alpha}^{2} + {\beta}^{2} = {3}^{2} + {7}^{2} = 9 + 49 = 58$