JEE Main 2023 — Mathematics Algebra

Given,

Quadratic equation ${x}^{2}-5\alpha x+1=0$x2−5αx+1=0 and ${x}^{2}-\alpha x-5=0$x2−αx−5=0 have a common real root,

So, taking $k$k to be common root then equation becomes,

$${k}^{2}-5\alpha k+1=0......(1)$$k2−5αk+1=0......(1)

$${k}^{2}-\alpha k-5=0.....(2)$$k2−αk−5=0.....(2)

Now subtraction above equations we get,

$$k=\frac{6}{4\alpha }$$k=4α6​

Now putting the value of $k$k in equation $(1)$(1) we get,

$${(\frac{6}{4\alpha })}^{2}-5\alpha (\frac{6}{4\alpha })+1=0$$(4α6​)2−5α(4α6​)+1=0

$$\Rightarrow {(\frac{6}{4\alpha })}^{2}-(\frac{26}{4})=0$$⇒(4α6​)2−(426​)=0

$$\Rightarrow \frac{36}{16{\alpha }^{2}}=(\frac{26}{4})$$⇒16α236​=(426​)

$$\Rightarrow {\alpha }^{2}=\frac{9}{26}$$⇒α2=269​

$$\Rightarrow \alpha =\frac{3}{\sqrt{2\times 13}}$$⇒α=2×13​3​

Now comparing with given value $\frac{3}{\sqrt{2\beta }}$2β​3​ we get, $\beta =13$β=13