If $x^{4} - 17 x^{2} + 16 = 0$ then find the sum of the squares of the roots

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

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

289

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

Open in App

Solution

The correct option is C 34
Let $x^{2} = a$
So, the given eqn becomes $a^{2} - 17 a + 16 = 0$
$=> a^{2} - 16 a - a + 16 = 0$
$=> a (a - 16) - 1 (a - 16) = 0$
$=> (a - 1) (a - 16) = 0$
$=> a = 1; 16$
So, $x^{2} = 1; 16$
$=> x = \pm 1; \pm 4$
So, Sum of squares of roots $= 1^{2} + {(- 1)}^{2} + 4^{2} + {(- 4)}^{2} = 1 + 1 + 16 + 16 = 34$

Suggest Corrections

Similar questions

x^{4} + 6 x^{3} + 17 x^{2} + 24 x + 16

x^{2} + 3 x + 4

f (x) = x^{4} - 12 x^{3} + 17 x^{2} - 9 x + 7

f (x + 3)

f (x) = x^{4} - 12 x^{3} + 17 x^{2} - 9 x - 7,

(1 - x)^{4} - 16 = 0

If x is multiplied with the sum of 11 x^{2} - 8 x + 4

and 6 x^{2} + 7 x - 10, then find the product.

Join BYJU'S Learning Program