Algebra Problem!

Question

Solve for x: 4x² - 3x = -12

Answer 1

The easiest way to solve this type of problem is to set it equal to zero so we can use the quadratic formula. Lets start by adding 12 to each side:

4x² - 3x + 12 = 0

Then we can apply the quadratic formula: x = [-b +/- √(b²-4*a*c)]/2*a

a = 4

b = -3

c = 12

so:

x =[3 +/- √(3²-4*4*12)]/2*4

x = [3 +/- √(9-192)]/8

x = [3 +/- √(-183)]/8

Now we have 2 equations:

x = 3/8 + [√(-183)]/8

x = 3/8 - [√(-183)]/8

The square root of a negative is the square root of the positive multiplied by i (√-1). We then divide by 8:

x = .375 + 1.69i

x = .375 - 1.69i

This is the furthest we can go. We can write it as x = .375 +/- 1.69i

Answer 2

Hi Jesset,

4x² - 3x = -12 both side +12

4x² - 3x +12=0. a=4. b= -3 C= 12

use quadratic formula

first find b² -4ac = 9 -4(4)(12)= -183<0 so this equation Has no real answer.

quadratic formula is:

x= (-b±√ b²-4ac)/2a

i hope it is useful,

Minoo

Answer 3

We need to isolate x, which is a bit tough when there's an exponent on x.

First, we move everything to one side, and get

4x² - 3x + 12 = 0

We can't solve this by factoring, and thus need to use the quadratic formula

The quadratic formula says:

x = [-b - sqrt (b²-4ac)]/(2a) and x = [-b + sqrt (b²-4ac)]/(2a)

In this case, we get x = 3/8 - sqrt(-183)/8 and x = 3/8 + sqrt(-183)/8

Since we can't take the square root of a negative, there are no real solutions.

However, if we want non-real solutions, where i = sqrt(-1), then we can get:

x = 3/8 - i*sqrt(183)/8 and x = 3/8 + i*sqrt(183)/8

Hope this helps!