Question

If a sq +1/a sq =47 and a not = 0 find. a+1/a

Answer 1

Given:

• $\sf{a^{2}+\frac{1}{a^{2}}=47}$

To Find:

• The value of $\sf{a+\dfrac{1}{a}}$

Concept Used:

• We will suppose that a+1/a = some variable ,then we will simplify and hence find the required answer.

Solution:.

We are given that $\sf{a^{2}+\dfrac{1}{a^{2}}=47}$ ..........(1)

Now , let $\sf{a+\dfrac{1}{a}=x}$

So , let's simplify :

$\sf{\implies a+\dfrac{1}{a}=x}$

On squaring both sides ,

$\sf{\implies (a+\dfrac{1}{a})^{2}=x^{2}}$

$\sf{\implies a^{2}+(\dfrac{1}{a})^{2}+2\times a\times \dfrac{1}{a}=x^{2}}$

using

• (a+b)²=a²+b²+2ab.

$\sf{\implies a^{2}+(\dfrac{1}{a})^{2}+2 =x^{2}}$

$\sf{\implies 47+2=x^{2}}$

[Putting value from (1) ]

$\sf{\implies x^{2}=49}$

$\sf{\implies x =\sqrt{49}}$

${\underline{\boxed{\red{\sf{\mapsto x =\pm7}}}}}$

Earlier we had supposed that $\sf{a+\dfrac{1}{a}=x}$ .

And , we got x = ±7

$\sf{\implies a+\dfrac{1}{a}=x}$

${\underline{\boxed{\red{\bf{\leadsto a+\dfrac{1}{a}=\pm7}}}}}$