If $a^2 +b^2 + 2b + 4a + 5 = 0$, then the value of $\frac{2a-3b}{2a+3b}$ is equal to:
Answer & explanation
Correct answer: option 1
If $a^2 +b^2 + 2b + 4a + 5 = 0$
We know that,
(a + b)2 = a2 + b2 + 2ab
a2 + b2 + 2b + 4a + 5 = 0,
= a2 + 4a + b2 + 2b + 5 = 0
= a2 + 4a + 4 + b2 + 2b + 1 = 0
= (a + 2)2 + (b + 1)2 = 0
So, a + 2 = 0
= a = -2
And, b + 1 = 0
= b = -1
Put these values in the desired equation,
$\frac{2a-3b}{2a+3b}$ = $\frac{2(-2)-3(-1)}{2(-2)+3(-1)}$ = \(\frac{1}{7}\)