4-2.Quadratic Equations and Inequations
hard

જો $a+b+c=1, a b+b c+c a=2$ અને $a b c=3$ હોય તો $a^{4}+b^{4}+c^{4}$ ની કિમંત મેળવો.

A

$15$

B

$13$

C

$17$

D

$21$

(JEE MAIN-2021)

Solution

${a^{2}+b^{2}+c^{2}=(a+b+c)^{2}-2 \Sigma a b=-3}$

${(a b+b c+c a)^{2}=\Sigma(a b)^{2}+2 a b c \sum a}$

${\Rightarrow \Sigma(a b)^{2}=-2}$

${a^{4}+b^{4}+c^{4}=\left(a^{2}+b^{2}+c^{2}\right)^{2}-2 \Sigma(a b)^{2}}$

${\quad=9-2(-2)=13}$

Standard 11
Mathematics

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