Question

Complete the square to find the minimum value of the expression 4x^2+8x+23

Answer 1

Alright! so how do we complete the square? you're first step is to take the coefficient in front of your middle value, or 8x. 8 would we the coefficent. now, you divide that value by 2.

so we have :

8/ 2 = 4

so what do we do with 4? we square it.

4^2 = 16

okay. so I've made you do some stuff. what do we do now? we add AND SUBTRACT this value from our equation. since I both add and subtract, I haven't changed the value of my equation.

4x^2 + 8x + 23 +16 -16

now, we isolate the positive version of the number we solved for--or 16, and all the terms that have a varaible. by this I mean:

(4x^2 + 8x + 16) + 23 - 16

all I've done is move stuff around. no values have been changed. so what now? well, you factor that portion in the parenthesis.

if you don't know how to factor, I can go over r that seperately, but I'm gonna assume you can factor. when I factor this, I get:

4x^2 + 8x + 16 = (x+ 4)(x + 4) = (x + 4)^2

so now I have

(x+4)^2 + 23 - 16

I can now combine like terms, which include 23-16, which equals 7.

finally, I have:

(x+4)^2 + 7

so I have officially completed the square. ^^^that's your process of doing that. so how do I find the minimum value? well, run with me for a sec. if I just look at (x+4)^2, what type of values will I always get? well, anything squared is POSITIVE, so I will Never get a negative number. in fact, I only get numbers from 0 to infinity. so, if the lowest number I can get there is 0, the lowest value this polynomial can have overall is 0+ 7, which is 7. hope that helps!