8X^2 + 2X - 1 = 0 need quadratic formula -b +/- sqrt(b^2-4ac)/2a a = 8 b = 2 c = -1 -2 +/- sqrt(2^2 -4(8)(-1))/2(8) -2 +/- sqrt(4 + 32)/16 -2 +/- sqrt(36)/16 (-2 +/- 6)/16 X = 1/4 X = -1/2 My TI-84 confirms this answer
(x + 2)(x - 7)
x2(x - 8)
80
That doesn't factor neatly. Applying the quadratic formula, we find two real solutions: (8 plus or minus the square root of 10) divided by 3 x = 3.720759220056127 x = 1.6125741132772067
8x2 + 2x - 1 = 8x2 + 4x - 2x - 1 = 4x(2x + 1) - 1(2x + 1) = (4x - 1)(2x + 1)
8X^2 + 2X - 1 = 0 need quadratic formula -b +/- sqrt(b^2-4ac)/2a a = 8 b = 2 c = -1 -2 +/- sqrt(2^2 -4(8)(-1))/2(8) -2 +/- sqrt(4 + 32)/16 -2 +/- sqrt(36)/16 (-2 +/- 6)/16 X = 1/4 X = -1/2 My TI-84 confirms this answer
What are the factors? 2x3 - 8x2 + 6x = 2x(x - 1)(x - 3).
8x2 - 12x - 8 = 4 (2x2 - 3x - 2) = 4 (2x2 + x - 4x - 2) = 4 (x (2x + 1) - 2 (2x + 1)) = 4 (x - 2) (2x + 1).
Simplifying 8x2 + -8x + -16 Factor out the Greatest Common Factor (GCF), '8'. 8(x2 + -1x + -2) Factor a trinomial. 8((-x + -1)(-x + 2)) Final result: 8(x - 2)(x + 1)
-8x2 - 2x + 8 this is a quadratic equation or a second order polynomial it is a second order polynomial because it has a term in x2 For every polynomial we name it according to the highest power term in the equation.......
(4x - 1)(2x + 1)
They are: (4x-5)(2x+5)
You add compatible terms together. For example, if one equation has 3x2, and the other has 5x2, you combine those two terms as 8x2.
-4x2 - 32x + x + 8 = 8x2 - 56x : simplify and group the various terms-32x + 56x + x + 8 = 8x2 + 4x225x + 8 = 12x2 : Now present this as a standard quadratic equation.12x2 - 25x - 8 = 0 : Solving for the roots of a quadratic equation gives :-x = [25 ± √(252 + 4x12x8)] ÷ 2x12 = [25 ± √1009] ÷ 24 = [25 ± 31.765] ÷ 24x = 56.765/24 = 2.365 and x = -6.765/24 = -0.282x = 2.365 and -0.282
(x + 2)(x - 7)
8x2 + 6x - 9 = 8x2 + 12x - 6x - 9 = 4x(2x + 3) - 3(2x + 3) = (2x + 3)(4x - 3)