Solve. Leave in simplest radical form.
1. x^2 + 8x = -16
2. 3x^2 = 2x + 5
Please help me solve. Thanks!
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
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