JEE MAIN - Mathematics (2023 - 30th January Evening Shift - No. 17)

If the value of real number $a>0$ for which $x^2-5 a x+1=0$ and $x^2-a x-5=0$

have a common real root is $\frac{3}{\sqrt{2 \beta}}$ then $\beta$ is equal to ___________.
Answer
13

Explanation

$${x^2} - 5\alpha x + 1 = 0$$ ..... (1)

$${x^2} - \alpha x - 5 = 0$$ ...... (2)

have a common root.

Subtracting (1) with (2) we'll get $$x = {6 \over {4\alpha }}$$

Substituting in (1)

$${{36} \over {16{\alpha ^2}}} - {{30} \over 4} + 1 = 0$$

$$ \Rightarrow {\alpha ^2} = {9 \over {26}}$$

$$\alpha = {3 \over {\sqrt {2 \times 13} }}$$

$$\therefore$$ $$\beta = 13$$

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