Thread: Solve

Solve. Leave in simplest radical form.

1. x^2 + 8x = -16

2. 3x^2 = 2x + 5

Please help me solve. Thanks!

Originally Posted by alwaysalillost

Solve. Leave in simplest radical form.

1. x^2 + 8x = -16

I presume you aren't allowed to use the quadratic formula. So I'll solve this using the "complete the square" method.

x^2 + 8x = -16

The form of a perfect square is
x^2 + (2a)x + a^2 = (x + a)^2

Comparing the LHS of both expressions (ie. comparing the coefficients of the x terms) I get that
2a = 8

a = 4

So we want to add a^2 = 4^2 = 16 to both sides of the equation. Thus
x^2 + 8x + 16 = -16 + 16

(x + 4)^2 = 0

x + 4 = 0

x = -4

-Dan

Thanks Topsquark!

Originally Posted by alwaysalillost

2. 3x^2 = 2x + 5

This one's a bit tougher:

3x^2 - 2x = 5

Divide through by the 3:
x^2 - (2/3)x = (5/3)

Now complete the square:
2a = -(2/3) ==> a = -(1/3)

So we need to add a^2 = 1/9 to both sides:
x^2 - (2/3)x + (1/9) = (5/3) + (1/9)

(x - 1/3)^2 = 16/9

x - 1/3 = (+/-)sqrt(16/9) = (+/-)(4/3)

x = (1/3) (+/-) (4/3)

So
x = 1/3 + 4/3 = 5/3
or
x = 1/3 - 4/3 = -1

-Dan

Thanks topsquark!