If 2a - 1-2a - 8- find - i 4a2 - 1-4a2 ii 16a4 - 1-16a4

(i) 4a^2 + 1/4a^2 = (2a + 1/2a)^2 2.2a 1/2a (2a +1/2a)^2 2 = (8)^2 2 = 64 2 = 62 (ii) 16a^4 + 1/16a^4 = (4a^2 + 1/4a^2)^2 2.4a^2.1/4a^2 =(62)^2 2 = ...

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

Byju's Answer

Standard VIII

Mathematics

Expansion of (x+y)^2

If 2a + 1/2a ...

Question

If $2 a + \frac{1}{2 a} = 8$ , find :

(i) $4 a^{2} + \frac{1}{4 a^{2}}$

(ii) $16 a^{4} + \frac{1}{16 a^{4}}$

Open in App

Solution

(i) $4 a^{2} + \frac{1}{4 a^{2}} = {(2 a + \frac{1}{2 a})}^{2} - 2.2 a \frac{1}{2 a} {(2 a + \frac{1}{2 a})}^{2} - 2 = (8)^{2} - 2 = 64 - 2 = 62$

(ii) $16 a^{4} + \frac{1}{16 a^{4}} = {(4 a^{2} + \frac{1}{4 a^{2}})}^{2} - 2.4 a^{2} . \frac{1}{4 a^{2}} = (62)^{2} - 2 = 3844 - 2 = 3842$

Suggest Corrections

Similar questions

If $2 a + \frac{1}{2 a} = 3$ , then $16 a^{4} + \frac{1}{16 a^{4}} = ?$

Q. If

2 a + \frac{1}{2 a} = 8

, then find

16 a^{4} + \frac{1}{16 a^{4}}

Q. If

2 a - \frac{1}{2 a} = 3

, then

16 a^{4} + \frac{1}{16 a^{4}}

is equal to:

Q. Evaluate:

(2 a - b) (2 a + b) (4 a^{2} + b^{2})

16 a^{4} - b^{4}

.
If true, then enter

1

and if false, then enter

0

Q. Find three geometric means between 1 and 16

a^{4}

Join BYJU'S Learning Program

If 2a+12a=8, find : (i) 4a2+14a2 (ii) 16a4+116a4

(i) 4a2+14a2=(2a+12a)2−2.2a12a(2a+12a)2−2=(8)2−2=64−2=62 (ii) 16a4+116a4=(4a2+14a2)2−2.4a2.14a2=(62)2−2=3844−2=3842

If $2 a + \frac{1}{2 a} = 8$ , find :

(i) $4 a^{2} + \frac{1}{4 a^{2}}$

(ii) $16 a^{4} + \frac{1}{16 a^{4}}$

(i) $4 a^{2} + \frac{1}{4 a^{2}} = {(2 a + \frac{1}{2 a})}^{2} - 2.2 a \frac{1}{2 a} {(2 a + \frac{1}{2 a})}^{2} - 2 = (8)^{2} - 2 = 64 - 2 = 62$

(ii) $16 a^{4} + \frac{1}{16 a^{4}} = {(4 a^{2} + \frac{1}{4 a^{2}})}^{2} - 2.4 a^{2} . \frac{1}{4 a^{2}} = (62)^{2} - 2 = 3844 - 2 = 3842$