Question

(a^2+1/a^2)=27 find (a+1/a) it is a question from expansions chapter class 9

Answer 1

Answer:

√29.

Step-by-step explanation:

Given,

     a^2 + 1 / a^2 = 27

Adding on both sides :

⇒ a^2 + 1 / a^2 + 2 = 27 + 2

⇒ a^2 + 1 / a^2 + 2( a x 1 / a ) = 29

Using a^2 + b^2 + 2ab = ( a + b )^2

⇒ ( a + 1 / a )^2 = 29

⇒ a + 1 / a = √29

Hence the required value of √29, however if you wanted the value of a - 1 / a answer could be 5 or - 5.

Answer 2

$\Large{\underline{\underline{\mathfrak{\bf{\pink{Question}}}}}}$

If (a²+1/a²)=27 find (a+1/a) = ?

$\Large{\underline{\underline{\mathfrak{\bf{\pink{Solution}}}}}}$

$\Large{\underline{\mathfrak{\bf{\red{Given}}}}}$

$\mapsto\sf{\green{\:(a^2+\dfrac{1}{a^2})\:=\:27}}$

$\Large{\underline{\mathfrak{\bf{\red{Find}}}}}$

$\mapsto\sf{\green{\:(a+\dfrac{1}{a})\:=\:?}}$

$\Large{\underline{\underline{\mathfrak{\bf{\pink{Explanation}}}}}}$

we know,

$\bigstar\sf{\green{\:(x+y)^2\:=\:(x^2+y^2+2xy)}}$

So,

$\mapsto\sf{\green{\:(a+\dfrac{1}{a})^2\:=\:(a^2+\dfrac{1}{a^2}+2.\red{\cancel{a}}.\dfrac{1}{\red{\cancel{a}}})}} \\ \\ \mapsto\sf{\:(a+\dfrac{1}{a})\:=\:\sqrt{(a^2+\dfrac{1}{a^2})+2}} \\ \\ \mapsto\sf{\:(a+\dfrac{1}{a})\:=\:\sqrt{(27)+2}} \\ \\ \mapsto\sf{\pink{\:(a+\dfrac{1}{a})\:=\:\sqrt{29}}}$

$\Large{\underline{\mathfrak{\bf{\red{Thus}}}}}$

$\mapsto{\red{\boxed{\sf{\orange{\:Value\:of\:(a+\dfrac{1}{a})\:=\:\sqrt{29}}}}}}$

Some Important Formula

★(a+b)^2=(a^2+b^2+2ab)

★(a-b)^2 = (a^2+b^2-2ab)

★(a-b)(a+b)= (a^2-b^2)

★(a+b)^3 = a^3+b^3+3ab(a+b)

★(a-b)^3 = a^3-b^3+3ab(a-b)

★(a^3-b^3)=(a-b)(a^2+b^2+ab)

★(a^3+b^3)=(a-b)(a^2+b^2-ab)