If x-1-v-20- then the value of x 2-1- x 2 isA- 395B- 405C- 402D- 398

The correct option is D 398Given that, x +1/x = 20 Squaring on both sides, ⇒ (x +1/x)^2 = (20)^2 = 400 nbsp;Using the identity (a + b)^2 = a^2 + b^2 + 2ab, ...

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

Byju's Answer

Standard VIII

Mathematics

Algebraic Quadratic Identity

If x+1/v=20, ...

Question

If $x + \frac{1}{x} = 20$ , then the value of
$x^{2} + \frac{1}{x^{2}}$ is _____.

395

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

405

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

402

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

398

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

Open in App

Solution

The correct option is D 398
Given that,
$x + \frac{1}{x} = 20$
Squaring on both sides,
$\Rightarrow (x + \frac{1}{x})^{2} = (20)^{2} = 400$
Using the identity $(a + b)^{2} = a^{2} + b^{2} + 2 a b$ ,
$x^{2} + \frac{1}{x^{2}} + 2 \times x \times \frac{1}{x} = 400$
$\Rightarrow x^{2} + \frac{1}{x^{2}} + 2 = 400$
$\Rightarrow x^{2} + \frac{1}{x^{2}} = 400 - 2 = 398$

Suggest Corrections

Similar questions

Q. If

x + \frac{1}{x} = 20

, then the value of

x^{2} + \frac{1}{x^{2}}

is .

Q. If

x + \frac{1}{x} = 5,

then the value

x^{2} + \frac{1}{x^{2}}

$I f x + \frac{1}{x} = 20, find the value of x^{2} + \frac{1}{x^{2}} .$

If x = ( $\sqrt$ 2+1)/( $\sqrt$ 2-1), and x -- y = 4 $\sqrt$ 2, then the value of (x $^{2}$ +y $^{2}$ )/2 is

Q. If

x^{2} + y^{2} + \frac{1}{x^{2}} + \frac{1}{y^{2}} = 4,

then the value of

x^{2} + y^{2}

Join BYJU'S Learning Program

If x+1x=20, then the value of x2+1x2 is _____.

The correct option is D 398Given that, x+1x=20 Squaring on both sides, ⇒(x+1x)2=(20)2=400 Using the identity (a+b)2=a2+b2+2ab, x2+1x2+2×x×1x=400 ⇒x2+1x2+2=400 ​​⇒x2+1x2=400−2=398

If $x + \frac{1}{x} = 20$ , then the value of
$x^{2} + \frac{1}{x^{2}}$ is _____.