If $9\left(a^2+b^2\right)+c^2+20=12(a+2 b)$, then the value of $\sqrt{6 a+9 b+2 c}$ is :
Answer & explanation
Correct answer: option 1
We know that,
( a - b )2 = a2 + b2 - 2ab
( a + b )2 = a2 + b2 + 2ab
9(a2 + b2 ) + c2 + 20 = 12(a + 2b)
= 9a2 + 9b2 + c2 + 20 = 12a + 24b
= 9a2 - 12a + 9b2 - 24b + c2 + 20 = 0
= 9a2 - 12a + 4 + 9b2 - 24b + 16 + c2 = 0
= (3a - 2)2 + (3b - 4)2 + c2 = 0
3a – 2 = 0
a = \(\frac{2}{3}\)
3b - 4 = 0
b = \(\frac{4}{3}\)
c = 0
$\sqrt{6 a+9 b+2 c}$ = $\sqrt{6 (\frac{2}{3}) +9 (\frac{4}{3})+2 (0)}$
$\sqrt{6 a+9 b+2 c}$ = \(\sqrt {16}\) = 4