Solve the quadratic equation 8x^2+6x-9
factors of 8 ... 2 and 4
factors of -9 ... 3 and -3
2 * -3 = -6 ... 4 * 3 = 12 ... -6 + 12 = 6
(2x + 3)(4x - 3) = 8x^2+6x-9
this is the factoring of the EXPRESSION
... an equation has an equal sign (=) in it
i got till (4x-3)(2x+3) bt how do u get x=3/4 or -3/2
As R_scott pointed out to you, it should have been
8x^2+6x-9 = 0
which led to
(4x-3)(2x+3) = 0
Now, how can the multiplication of 2 numbers end up being zero???
Only if one of them is zero, but which one?
Well, it could have been either one, that is
4x-3 = 0 OR 2x+3 = 0
now just solve those two little equations
4x-3 = 0
4x = 3
x = 3/4 , same for the other one
To solve the quadratic equation 8x^2 + 6x - 9, we can use the quadratic formula. The quadratic formula states that for an equation in the form of ax^2 + bx + c = 0, the solutions for x can be found using the formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
For the given equation 8x^2 + 6x - 9, we can identify a = 8, b = 6, and c = -9. Plugging these values into the quadratic formula, we get:
x = (-6 ± √(6^2 - 4 * 8 * -9)) / (2 * 8)
Now, let's simplify the equation within the square root:
x = (-6 ± √(36 + 288)) / 16
x = (-6 ± √324) / 16
x = (-6 ± 18) / 16
We have two possible solutions:
x1 = (-6 + 18) / 16 = 12/16 = 3/4
x2 = (-6 - 18) / 16 = -24/16 = -3/2
Therefore, the solutions to the quadratic equation 8x^2 + 6x - 9 are x = 3/4 a
nd x = -3/2.