Question

The quadratic 4x^2+2x-1 can be written in the form a(x+b)^2+c, where a, b, and c are constants. What is a+b+c?

Answer 1

`\large \boxed{\sf \ \ \ a+b+c=3 \ \ \ }`

Explanation:

Hello,

`4x^2+2x-1=4(x^2+(2)/(4)x)-1=4[(x+(1)/(4))^2-(1)/(4^2)]-1`

As

`(x+(1)/(4))^2=x^2+(2)/(4)x+(1^2)/(4^2)=x^2+(2)/(4)x+(1)/(4^2) \ \ So \\\\x^2+(2)/(4)x=(x+(1)/(4))^2-(1)/(4^2)`

Let 's go back to the first equation

`4x^2+2x-1=4[(x+(1)/(4))^2-(1)/(4^2)]-1=4(x+(1)/(4))^2-(1)/(4)-1\\\\=4(x+(1)/(4))^2-(1+4)/(4)=\boxed{4(x+(1)/(4))^2-(5)/(4)}`

a = 4

`b=(1)/(4)\\\\c=-(5)/(4)`

`a+b+c=4+(1)/(4)-(5)/(4)=(4*4+1-5)/(4)=(12)/(4)=3`

Hope this helps.

Do not hesitate if you need further explanation.

Thank you