Question

h(x)=x^2+10x+16 can anyone complete the square and explain the step?

Answer 1

whats the coefficient of x ?

Answer 2

10 right?

Answer 3

yes, take half of it : 10/2 = 5

Answer 4

square it,
then add and subtract

Answer 5

h(x) = x^2 + 10x + 16
= x^2 + 10x + 5^2 - 5^2 + 16

Answer 6

recall the identity : `a^2 + 2ab + b^2 = (a+b)^2`

Answer 7

h(x) = x^2 + 10x + 16
= x^2 + 10x + 5^2 - 5^2 + 16
= (x+5)^2 - 5^2 + 16

Answer 8

we're done ! just simplify the last two terms ^

Answer 9

So (x+5)^2 -9 is the answer?

I have to find 2 complex solutions, is this one?

Answer 10

thats `completing the square` part of the problem

Answer 11

to find the solutions, set it equal to 0 and solve x

Answer 12

From the complete the square part of from the beginning?

Answer 13

(x+5)^2 -9 = 0

(x+5)^2 = 9

x+5 = +-3

Answer 14

solve x^

Answer 15

So its just 2/-2?

Answer 16

x+5 = +-3

x+5 = 3 ; x+5 = -3
x = 3-5 ; x = -3-5
x = -2 ; x = -8

Answer 17

Ohhh, right, oops. But those arent complex solutions are they..? What kid of equation will give us one?

Answer 18

to get complex solutions, you need an equation with below condition :
b^2 - 4ac < 0

Answer 19

try below :

h(x) = x^2 + 2x + 2

Answer 20

So we'd have x^2+2x+SQRT2^2-SQRT2^2+2?

Answer 21

here are the steps to complete the square :
1) take half of x coefficient
2) square it, add and subtract
3) complete the square using the identity `a^2 + 2ab + b^2 = (a+b)^2`
4) simplify the tail

Answer 22

h(x) = x^2 + 2x + 2

1) take half of x coefficient
what do you get ?

Answer 23

1

Answer 24

2) square it, add and subtract

what do you get ?

Answer 25

x^2+2x+1^2-1^2

Answer 26

+2

Answer 27

don't forget the constant term 2 in the end ^

Answer 28

yes !

h(x) = x^2 + 2x + 2
= x^2 + 2x + 1^2 - 1^2 + 2

Answer 29

And then we have 1+4+4=(a+b)^2 When we do that A^2+2ab+b^2=(a+b)^2

Answer 30

yes, a = x, b = 1

Answer 31

h(x) = x^2 + 2x + 2
= x^2 + 2x + 1^2 - 1^2 + 2
= (x+1)^2 -1^2 + 2

Answer 32

simplify

Answer 33

So really, it is just (x+1)^2+1?

Answer 34

Yep ! set it equal to 0 and solve x

Answer 35

So what do I do after -1=(x+1)^2?

Answer 36

take sqrt both sides

Answer 37

And then 1i=x+1 goes to x=1i-1?

Answer 38

very good, but looks u made a small mistake.

\(\large (x+1)^2 = -1 \)

take sqrt both sides :
\(\large (x+1) = \pm \sqrt{-1} \)

\(\large x+1 = \pm i \)
\(\large x = ?\)

Answer 39

x=1+/- i?

Answer 40

-1*

Answer 41

Yes !

Answer 42

\(\large x = -1 \pm i\)

Answer 43

Thank you for the walkthrough, I really appreciate it!

Answer 44

np, you're welcome :)