Math Help PLEASE?

Question

1. x^2 + 4x =3
A. -2 -+ √7
B. 2 -+ 2 √7
C. -4 -+ 2 √7
D. no real number solution

2. 2x^2 + 4 = 5x
A. -2 +- √4 / 2
B. 5 +- √57/ 4
C. -1 +- 2 √14
D. no real number solution

Answer 1

1. x^2 + 4x =3
A. 2 -+ √7

[-b ± √(b² - 4ac)] / 2a
[-4 ± √ (16 - (-12)) ] / 2
[-4 ± √28] / 2
[-4 ± √(4)(7)] / 2
[-4 ± 2√7] / 2
[-2 ± √7]

2. 2x^2 + 4 = 5x
D. no real number solution

[-b ± √(b² - 4ac)] / 2a
[5 ± √(25 - (4)(2)(4))] / 4
[5 ± √(25 - 32)] / 4
[5 ± √(-7) ] / 4

A negative discriminant indicates no real roots

Answer 2

x^2 + 4x = 3
x^2 + 4x - 3 = 0

[-4 +/- sqrt(4^2 - 4(1)(-3))] / 2(1)
[-4 +/- sqrt(16 + 12)] / 2
[ -4 +/- sqrt(28)]/2
[-4 +/- sqrt(7*4)] / 2
-4 +/- 2sqrt(7)] / 2
-2 +/- sqrt(7)

2x^2 + 4 = 5x
2x^2 -5x + 4 = 0

The discriminate is 25 - 4(2)(4) = -7. So, there is no real solution.

Answer 3

Use the quadratic equation
x = (- b ± √(b^2 - 4ac))/2
Or you can factor by completing the square and set factors equal to 0.

Answer 4

the answer is 2

Answer 5

I love the homework questions. I'm going to start searching questions posted by my kids and catch 'em in the act.