Posted by **timmyj** on Tuesday, February 12, 2008 at 2:25pm.

1. solve: x^3 =4x

2. solve: (x-3)^2 = -4

3. solve: (6 +5i) ^2

4.Two integers have a sum of -4 The sum of their squares is 40. What are the two integers?

- algebra 2 -
**drwls**, Tuesday, February 12, 2008 at 2:43pm
1. x(x^2-4) = 0

x(x+2)(x-2) = 0

Do you see the three possibilties now?

2. Take the square root of both sides and then do some algebra to get x

3. There is nothing to solve. You do not have an equation.

4. Call the numbers x and y

x + y = -4, so y = -4-x

Then solve the equation

x^2 + (-4-x)^2 = 40

2x^2 +8x -24 = 0

x^2 +4x -12 = 0

(x+6)(x-2) = 0

- algebra 2 -
**Guido**, Tuesday, February 12, 2008 at 5:28pm
I will do number 2 only.

We have this:

(x-3)^2 = -4

The left side becomes x^2 - 6x + 9.

We now have this:

x^2 - 6x + 9 = -4

Add 4 to both sides and then equate to zero.

x^2 - 6x + 9 + 4 = 0

x^2 - 6x + 13 = 0

This quadratic equation cannot be solved by factoring. In that case, use the quadratic formula to find x.

If you do not know how to use the quadratic formula, write back.

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