Find the square root of 8+sqrt{63} in the for | vedclass

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(B) Let the square root be $\sqrt{a} + \sqrt{b}$.By squaring both sides,we get $a + b + 2\sqrt{ab} = 8 + \sqrt{63}$.Comparing the rational and irratio

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$\frac{3\sqrt{2} + \sqrt{14}}{2}$

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(B) Let the square root be $\sqrt{a} + \sqrt{b}$.By squaring both sides,we get $a + b + 2\sqrt{ab} = 8 + \sqrt{63}$.Comparing the rational and irrational parts,we have $a + b = 8$ and $2\sqrt{ab} = \sqrt{63}$.Squaring the second equation,$4ab = 63$,so $ab = \frac{63}{4}$.Now,consider a quadratic equation with roots $a$ and $b$: $t^2 - (a+b)t + ab = 0$,which is $t^2 - 8t + \frac{63}{4} = 0$.Multiplying by $4$,we get $4t^2 - 32t + 63 = 0$.Using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $t = \frac{32 \pm \sqrt{1024 - 1008}}{8} = \frac{32 \pm 4}{8}$.Thus,$t_1 = \frac{36}{8} = \frac{9}{2}$ and $t_2 = \frac{28}{8} = \frac{7}{2}$.The square root is $\sqrt{\frac{9}{2}} + \sqrt{\frac{7}{2}} = \frac{3}{\sqrt{2}} + \frac{\sqrt{7}}{\sqrt{2}} = \frac{3 + \sqrt{7}}{\sqrt{2}}$.Rationalizing the denominator by multiplying the numerator and denominator by $\sqrt{2}$,we get $\frac{(3 + \sqrt{7})\sqrt{2}}{2} = \frac{3\sqrt{2} + \sqrt{14}}{2}$.

Find the square root of $8+\sqrt{63}$ in the form of a binomial surd.

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