Question

If (3a + 1)2 + (b − 1)2 + (2c − 3)3 = 0, then the value of (3a+b+2c) is equal to: ​

Answer 1

Answer:

(3a+1)²+(b−1)²+(2c-3)³=0

(3a+1)²=0

3a=-1

a= -1/3

(b-1)²=0

b-1=0

b=1

(2c-3)³=0

c= 3/2

∴3a+b+2c (We need to evaluate this)

=3×(-1/3)+1+3/2×2 (Substituting the values we just found)

= -1+1+3

=3

The final answer=3.

Hope this helps!