Question

a+ 1/a = 34 then✓ a +1/✓ a = ?​

Answer 1

GIVEN :-

• $\rm{a+\dfrac{1}{a}= 34}$

TO FIND :-

• $\rm{\sqrt{a} +\dfrac{1}{\sqrt{a}}}$

IDENTITY USED:-

• ${\boxed{\blue{\red{ (a+b)^2= a^2+b^2+2ab}}}}$

Now,

$\implies\rm{(\sqrt{a}+\dfrac{1}{\sqrt{a}})^2= (\sqrt{a})^2+(\dfrac{1}{\sqrt{a}})^2+2\times{\sqrt{a}}\times{\dfrac{1}{\sqrt{a}}}}$

$\implies\rm{(\sqrt{a}+\dfrac{1}{\sqrt{a}})^2= a+\dfrac{1}{a} + 2}$

$\implies\rm{(\sqrt{a}+\dfrac{1}{\sqrt{a}})^2 = 34 +2}$

$\implies\rm{(\sqrt{a}+\dfrac{1}{\sqrt{a}})^2 = 36}$

$\implies\rm{(\sqrt{a}+\dfrac{1}{\sqrt{a}})^2 = \sqrt{36}}$

$\implies\rm{\sqrt{a}+\dfrac{1}{\sqrt{a}} = 6}$

Hence , The value of $\rm{\sqrt{a} +\dfrac{1}{\sqrt{a}}}$ is 6.

$\boxed{\begin{minipage}{5cm}\\ \\ \sf\bf{(a-b)^2 = \tt a^2+b^2-2ab }\\ \\ \tt\bf{a^3+b^3= \tt (a+b) (a^2+ab+b^2) }\\ \\ \rm\bf{a^3-b^3= \tt (a-b) (a^2+ab+b^2) }\end{minipage}}$