Solve for x
x=-2
x = \frac{4}{3} = 1\frac{1}{3} \approx 1.333333333
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a+b=2 ab=3\left(-8\right)=-24
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx-8. To find a and b, set up a system to be solved.
-1,24 -2,12 -3,8 -4,6
Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -24.
-1+24=23 -2+12=10 -3+8=5 -4+6=2
Calculate the sum for each pair.
a=-4 b=6
The solution is the pair that gives sum 2.
\left(3x^{2}-4x\right)+\left(6x-8\right)
Rewrite 3x^{2}+2x-8 as \left(3x^{2}-4x\right)+\left(6x-8\right).
x\left(3x-4\right)+2\left(3x-4\right)
Factor out x in the first and 2 in the second group.
\left(3x-4\right)\left(x+2\right)
Factor out common term 3x-4 by using distributive property.
x=\frac{4}{3} x=-2
To find equation solutions, solve 3x-4=0 and x+2=0.
3x^{2}+2x-8=0
All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.
x=\frac{-2±\sqrt{2^{2}-4\times 3\left(-8\right)}}{2\times 3}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 2 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-2±\sqrt{4-4\times 3\left(-8\right)}}{2\times 3}
Square 2.
x=\frac{-2±\sqrt{4-12\left(-8\right)}}{2\times 3}
Multiply -4 times 3.
x=\frac{-2±\sqrt{4+96}}{2\times 3}
Multiply -12 times -8.
x=\frac{-2±\sqrt{100}}{2\times 3}
Add 4 to 96.
x=\frac{-2±10}{2\times 3}
Take the square root of 100.
x=\frac{-2±10}{6}
Multiply 2 times 3.
x=\frac{8}{6}
Now solve the equation x=\frac{-2±10}{6} when ± is plus. Add -2 to 10.
x=\frac{4}{3}
Reduce the fraction \frac{8}{6} to lowest terms by extracting and canceling out 2.
x=-\frac{12}{6}
Now solve the equation x=\frac{-2±10}{6} when ± is minus. Subtract 10 from -2.
x=-2
Divide -12 by 6.
x=\frac{4}{3} x=-2
The equation is now solved.
3x^{2}+2x-8=0
Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.
3x^{2}+2x-8-\left(-8\right)=-\left(-8\right)
Add 8 to both sides of the equation.
3x^{2}+2x=-\left(-8\right)
Subtracting -8 from itself leaves 0.
3x^{2}+2x=8
Subtract -8 from 0.
\frac{3x^{2}+2x}{3}=\frac{8}{3}
Divide both sides by 3.
x^{2}+\frac{2}{3}x=\frac{8}{3}
Dividing by 3 undoes the multiplication by 3.
x^{2}+\frac{2}{3}x+\left(\frac{1}{3}\right)^{2}=\frac{8}{3}+\left(\frac{1}{3}\right)^{2}
Divide \frac{2}{3}, the coefficient of the x term, by 2 to get \frac{1}{3}. Then add the square of \frac{1}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+\frac{2}{3}x+\frac{1}{9}=\frac{8}{3}+\frac{1}{9}
Square \frac{1}{3} by squaring both the numerator and the denominator of the fraction.
x^{2}+\frac{2}{3}x+\frac{1}{9}=\frac{25}{9}
Add \frac{8}{3} to \frac{1}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x+\frac{1}{3}\right)^{2}=\frac{25}{9}
Factor x^{2}+\frac{2}{3}x+\frac{1}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.
\sqrt{\left(x+\frac{1}{3}\right)^{2}}=\sqrt{\frac{25}{9}}
Take the square root of both sides of the equation.
x+\frac{1}{3}=\frac{5}{3} x+\frac{1}{3}=-\frac{5}{3}
Simplify.
x=\frac{4}{3} x=-2
Subtract \frac{1}{3} from both sides of the equation.
x ^ 2 +\frac{2}{3}x -\frac{8}{3} = 0
Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.This is achieved by dividing both sides of the equation by 3
r + s = -\frac{2}{3} rs = -\frac{8}{3}
Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C
r = -\frac{1}{3} - u s = -\frac{1}{3} + u
Two numbers r and s sum up to -\frac{2}{3} exactly when the average of the two numbers is \frac{1}{2}*-\frac{2}{3} = -\frac{1}{3}. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>
(-\frac{1}{3} - u) (-\frac{1}{3} + u) = -\frac{8}{3}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{8}{3}
\frac{1}{9} - u^2 = -\frac{8}{3}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{8}{3}-\frac{1}{9} = -\frac{25}{9}
Simplify the expression by subtracting \frac{1}{9} on both sides
u^2 = \frac{25}{9} u = \pm\sqrt{\frac{25}{9}} = \pm \frac{5}{3}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-\frac{1}{3} - \frac{5}{3} = -2 s = -\frac{1}{3} + \frac{5}{3} = 1.333
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
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