Question

x2 + 8x -16= 0 solve quadratic equation using one of the 4 methods.

Answer 1

i tried the completing the square and it dosent match any answers i have.. ive done it several times.

Answer 2

I have it finally

Answer 3

D=b^2-4ac
=64+4*16
=64+64=128
so you can find x1 and x2

Answer 4

That's factoring by grouping

Answer 5

i did that and got the same thing idk why it dosent have the answer on it , ahh! lol

Answer 6

x^2 + 8x -16 = 0
x^2 + (4+4sqrt{2})x + (4-4sqrt{2})x -16 = 0
x(x+ 4 + 4sqrt{2})+(4-4sqrt{2})(x+4+4sqrt{2}) = 0
(x+4+4sqrt{2})(x+4-4sqrt{2}) = 0

x = -4 - 4sqrt{2}
x = -4 +4sqrt{2}

Answer 7

There we go. Try that

Answer 8

another way of doing this is as follows:\[x^2+8x-16=x^2+8x+(-16)\]\[=x^2+8x+(16-32)\]\[=x^2+8x+16-32\]\[=(x^2+8x+16)-32\]\[=(x+4)^2-32\]then use this in your original equation to get:\[(x+4)^2-32=0\]\[(x+4)^2=32\]\[x+4=\pm\sqrt{32}=\pm\sqrt{16*2}=\pm4\sqrt{2}\]\[x=-4\pm4\sqrt{2}\]

Answer 9

this gives you the same answer as @Hero found

Answer 10

How'd you come up with the (16-32) part? What's the technique for that?

Answer 11

I think I know anyway, but...

Answer 12

I think Wiseman gave up

Answer 13

i usually concentrate on just the first two terms of the equation and ignore the rest.
so I would just look at:\[x^2+8x\]and ask myself what would give me these terms. the answer is:\[(x+4)^2\]I then expand that and see what I need to do to make the result match the remaining term.
Hope the explanation is clear.

Answer 14

Yeah, I figured as much. I get it

Answer 15

My method is messy

Answer 16

all roads lead to Rome :-)