These questions are pertaining to College Algebra

Question

Solve: 2x^2 + x - 15 = 0

 

Solve: (2 - 3x)^2 = 8

 

Solve: -4 x^2 = -12 x + 11

Answer 1

(1)

2x² + x - 15 = 0

Let's look at the pieces:

(2x ____) (x ____) = 0

That 2 is a prime number eases how the factoring works.

We also know we're going to have a +- in some fashion to account for that -15.

 

15 = 1 * 15 or 3 * 5

15 can be multiplied in two ways (if we're dealing with integers).

 

1 = 2 * 3 - 5

For that single x in the center, we need to add/subtract. The multiplying 2 comes from the 2 in the parentheses.

 

Now to put the pieces together:

(2x - 5)(x + 3) = 0

FOIL to check the answer.

That leaves us with 2x - 5 = 0 AND x + 3 = 0. You can solve those, I'm sure.

 

(2)

(2 - 3x)² = 8

Take the square root:

2 - 3x = ± √8

-3x = -2 ± √8

x = (-2 ± √8) / -3

x = (2 ± √8) / 3

 

(3)

-4x² = -12x + 11

0 = 4x² - 12x + 11

As we have a non-prime number next to the x², I'll be using the quadratic formula in this case:

When ax2 + bx + c = 0, x = (-b ± √(b² -4ac)) / 2a

x = (-(-12) ± √((-12)² - 4 * 4 * 11)) / (2 * 4)

x = (12 ± √(144 - 176)) / 8

x = (12 ± √-32) / 8

We cannot take the root of a negative, so there are no real solutions. But if you want, we can go to imaginary solutions instead:

x = (12 ± √(-1 * 32) / 8

x = (12 ± √-1 √(16 * 2) / 8

x = (12 ± 4i√2) / 8

x = (3 ± i√2) / 2