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4-2.Quadratic Equations and Inequations
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If $a, b, c$ are real numbers such that $a+b+c=0$ and $a^2+b^2+c^2=1$, then $(3 a+5 b-8 c)^2+(-8 a+3 b+5 c)^2$ $+(5 a-8 b+3 c)^2$ is equal to
A
$49$
B
$98$
C
$147$
D
$294$
(KVPY-2017)
Solution
(c)
We have, $a+b+c=0$ and $a^2+b^2+c^2=1$
Now $(3 a+5 b-8 c)^2+(-8 a+3 b+5 c)^2 +(5 a-8 b+3 c)^2$
$=9 a^2+25 b^2+64 c^2-48 a c+30 a b -80 b c+64 a^2+9 b^2+25 c^2-80 a c -48 a b+30 b c+25 a^2+64 b^2+9 c^2 +30 a c-8 a b-48 b c$
$=98\left(a^2+b^2+c^2\right)-98(a b+b c+c a)$
$=98\left(a^2+b^2+c^2\right) -98\left(\frac{(a+b+c)^2-\left(a^2+b^2+c^2\right)}{\text { 2 }}\right)$
$=98(1)-98\left(\frac{0-1}{2}\right)=98+49=147$
Standard 11
Mathematics
