Solve y=x+13 and y=|2x^2-8| Thanks?

This can still be solved with substitution.

y = x + 13 and y = |2x² - 8|

Both are equal to y, so:

x + 13 = |2x² - 8|

Removing the absolute value adds a ± to the other side:

±(x + 13) = 2x² - 8

simplify this into two separate equations to be solved:

x + 13 = 2x² - 8 and -x - 13 = 2x² - 8
0 = 2x² - x - 21 and 0 = 2x² + x + 5

The first one factors, but the second one has no real roots. The second equation's discriminant:

b² - 4ac
1² - 4(2)(5)
1 - 40
-39

So, no real roots, so we can throw that out.

That leaves us with:

0 = 2x² - x - 21
0 = (2x - 7)(x + 3)
x = -3 and 7/2

Now that we have two x's, we can find two y's:

y = x + 13
y = -3 + 13 and y = 7/2 + 13
y = 10 and y = 7/2 + 26/2
y = 10 and y = 33/2

so we have two points:

(-3, 10) and (7/2, 33/2)

x + 13 = |2x^2 - 8|

(-3, 10), (3.5, 16.5) <<< answers