algebra....anyone good at these types of problems.?

Question

1. The only way to solve for x on this equation: 2x^2 - 5x - 3 = 0; is to use the quadratic formula

true or false?

2, x^2 - 3x - 5 = 0

solve for x

3. x^2 - 5x - 2 = 0

4. Solve for x

3x^2 + 10x + 5 = 0

5. x^2 - 6x + 8 = 0

x = _______ and x = _______

6. 5x^2 + 2x - 3 = 0

7. 2x^2 - 12x = -18


x = ______

8. 3x^2 + 2x - 8 = 0

solve for x

9. 2x^2 + 7x - 4 = 0

10. 4x^2 - 36 = 0

means that x = +/- ______

Answer 1

1. False. 2x^2-5x-3 = (2x+1)(x-3)
2 True
3. x=[5 +/- sqrt(25+8)]/2 = 2.5 +/- .5sqrt*33)
4. x = [-10 +/-sqrt(100- 60)]/6
= -5/3 +/- (1/3)sqrt(10)
5. x+ 2 and x = 4
6. (5x-3)(x+1) --> x = 3/5 and -1
7. x = 3
8. Do the rest yourself.

Answer 2

1)true
2)true
3)x=5/2+sqrt 3/2 ,and, x=5/2-sqrt3/2
4)x=-10/6 +sqrt50i/6 ,and, x=-10/6 -sqrt50i/6
5)x=4 ,and, x=2
6)x=3/5 ,and, x=-1/5
7)x=1.5+sqrt156i/8,and, x=1.5-sqrt156i/8
8)x=2/3 ,and, x=-4/3
9)x=-7/4+sqrt39/4 ,and, x=-7/4-sqrt39/4
10)x=2/3 ,and, x=-2/3

Answer 3

Hey, Piggy, only ONE problem per question!!