Question

x^2-6x+8=0
x^2-4x-5=0
x^2+8x+15=0
-x^2-2x+3=0
x^2+4x+8=0
-x^2+2x=12
x^2-4x+2=0
x^2-6x+9=0

Answer 1

I have to graph the answer @thomaster

Answer 2

Do you need to solve for x?

Answer 3

yes @thomaster

Answer 4

okay the first one \(x^2-6x+8=0\)
You either need to get binomials in parentheses or use the quadratic formula.
The general form of these quadratic equations is \(ax^2+bx+c=0\)
To get the stuff in parentheses you need to find 2 numbers that are bx when added and c when multiplied.

For the example we need to find 2 numbers who are -6 when added to eachother and 8 when multiplied. It's just a little bit of trial and error. I can see that it has to do something with 4 and 2 because 4*2 = 8 and 4+2 = 6. the only problem is that its -6 and not 6. So you play a little bit with the numbers:
-4*2 = -8 and -4+2 = 2 so this is not correct
4*-2 = -8 (you can see it's incorrect already here because it needs to be 8)
-4*-2=8 and -4+-2=-6 (+- is the same as -)

Now you have the numbers and put them in parentheses like this:
\(\Large(x-4)(x-2)=0\)
you know this is correct by distributing the binomials again, you'd get \(x^2-6x+8=0\)
to continue:
\((x-4)(x-2)=0\)
\((x-4)=0~~and~~(x-2)=0\)
\(x-4=0~~and~~x-2=0\) first one you add 4 on both sides and 2 at the second
\(x=4~~and~~x=2\)

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