If $a, b, c, d$ are the roots of $x^{4} - 8 x^{3} + 8 x^{2} - 10 x + 5 = 0$ , then the value of $a^{2} + b^{2} + c^{2} + d^{2}$ , is:

8

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

48

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

56

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

80

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

Open in App

Solution

The correct option is D $48$
Given expression is $x^{4} - 8 x^{3} + 8 x^{2} - 10 x + 5 = 0$
We have,
$a^{2} + b^{2} + c^{2} + d^{2} = (a + b + c + d)^{2} - 2 (a b + a c + a d + b c + b d + a d)$
Now $a + b + c + d = \frac{8}{1}$
$a b + a c + a d + b c + b d + a d = \frac{8}{1}$
Hence, $a^{2} + b^{2} + c^{2} + d^{2} = (8)^{2} - 2 (8) = 64 - 16 = 48$

Suggest Corrections

Similar questions

a, b, c, d

x^{4} + 5 x^{3} - 4 x^{2} + 6 = 0

\frac{1}{a^{2} - 5 a + 6} + \frac{1}{b^{2} - 5 b + 6} + \frac{1}{c^{2} - 5 c + 6} + \frac{1}{d^{2} - 5 d + 6}

A^{2} + B^{2} + C^{2} + D^{2}

If a, b, c, d are in G.p., prove that :

(i) $(a^{2} + b^{2}), (b^{2} + c^{2}), (c^{2} + d)^{2}$ are in G.P.

(ii) $(a^{2} - b^{2}), (b^{2} - c^{2}), (c^{2} - d)^{2}$ are in G.P.

(iii) $\frac{1}{a^{2} + b^{2}}, \frac{1}{b^{2} + c^{2}}, \frac{1}{c^{2} + d^{2}}$ are in G.P.

(iv) $(a^{2} + b^{2} + c^{2}), (a b + b c + c d), (b^{2} + c^{2} + d^{2})$

a^{2} + b^{2} = c^{2} + d^{2}

a^{2} + d^{2} - a d = b^{2} + c^{2} + b c

\frac{a b + c d}{a d + b c}

a, b, c, d

\frac{a^{2} + b^{2} + c^{2}}{d^{2}}

Join BYJU'S Learning Program