If the value of a real number a0 for which x^2-5ax+1=0 and x^2-ax-5=0 have a common real …

Question

If the value of a real number $a>0$ for which $x^2-5ax+1=0$ and $x^2-ax-5=0$ have a common real root is $\dfrac{3}{\sqrt{2\beta}}$, then $\beta$ is equal to \_\_\_\_\_\_\_\_.

Answer: 13

Accepted Answer

Subtracting the equations: $-4ax+6=0 \implies x=\dfrac{3}{2a}$ (the common root).

Substituting into $x^2-ax-5=0$:
$$\frac{9}{4a^2}-\frac32-5=0 \implies \frac9{4a^2}=\frac{13}2 \implies a^2=\frac9{26} \implies a=\frac3{\sqrt{26}}.$$

Comparing with $a=\dfrac3{\sqrt{2\beta}}$: $2\beta=26 \implies \beta=13.$

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