The correct option is D (x+1/x 3)(x+1/x+3)x^2+1/x^2 11 Splitting 11 as the sum of ( 2) and ( 9), we have = (x^2+1/x^2 2 9) (x + 1/x)^2 = x^2 2 +1/x^2 = ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard IX

Mathematics

Perfect Square Trinomials

Factorise x2+...

Question

Factorise $x^{2} + \frac{1}{x^{2}} - 11$

(x + \frac{1}{x^{2}} - 3) (x + \frac{1}{x^{2}} + 3)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(x + \frac{1}{x^{2}} + 3) (x + \frac{1}{x} + 3)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(x^{2} + x - 3) (x + \frac{1}{x^{2}} + 3)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(x + \frac{1}{x} - 3) (x + \frac{1}{x} + 3)

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $(x + \frac{1}{x} - 3) (x + \frac{1}{x} + 3)$
$x^{2} + \frac{1}{x^{2}} - 11$
Splitting -11 as the sum of (-2) and (-9), we have
$= (x^{2} + \frac{1}{x^{2}} - 2 - 9)$ $(x + \frac{1}{x})^{2} = x^{2} - 2 + \frac{1}{x^{2}}$
$= (x + \frac{1}{x})^{2} - 9$
We know $a^{2} - b^{2} = (a + b) (a - b)$
$= (x + \frac{1}{x})^{2} - 3^{2}$
$= (x + \frac{1}{x} - 3) (x + \frac{1}{x} + 3)$

Suggest Corrections

Factorise x2+1x2−11

The correct option is D (x+1x−3)(x+1x+3)x2+1x2−11 Splitting -11 as the sum of (-2) and (-9), we have =(x2+1x2−2−9) (x+1x)2=x2−2+1x2 =(x+1x)2−9 We know a2−b2=(a+b)(a−b) =(x+1x)2−32 =(x+1x−3)(x+1x+3)

Factorise $x^{2} + \frac{1}{x^{2}} - 11$