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3x^{2}-x-2=0
Divide both sides by 2.
a+b=-1 ab=3\left(-2\right)=-6
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx-2. To find a and b, set up a system to be solved.
1,-6 2,-3
Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -6.
1-6=-5 2-3=-1
Calculate the sum for each pair.
a=-3 b=2
The solution is the pair that gives sum -1.
\left(3x^{2}-3x\right)+\left(2x-2\right)
Rewrite 3x^{2}-x-2 as \left(3x^{2}-3x\right)+\left(2x-2\right).
3x\left(x-1\right)+2\left(x-1\right)
Factor out 3x in the first and 2 in the second group.
\left(x-1\right)\left(3x+2\right)
Factor out common term x-1 by using distributive property.
x=1 x=-\frac{2}{3}
To find equation solutions, solve x-1=0 and 3x+2=0.
6x^{2}-2x-4=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{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 6\left(-4\right)}}{2\times 6}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, -2 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-2\right)±\sqrt{4-4\times 6\left(-4\right)}}{2\times 6}
Square -2.
x=\frac{-\left(-2\right)±\sqrt{4-24\left(-4\right)}}{2\times 6}
Multiply -4 times 6.
x=\frac{-\left(-2\right)±\sqrt{4+96}}{2\times 6}
Multiply -24 times -4.
x=\frac{-\left(-2\right)±\sqrt{100}}{2\times 6}
Add 4 to 96.
x=\frac{-\left(-2\right)±10}{2\times 6}
Take the square root of 100.
x=\frac{2±10}{2\times 6}
The opposite of -2 is 2.
x=\frac{2±10}{12}
Multiply 2 times 6.
x=\frac{12}{12}
Now solve the equation x=\frac{2±10}{12} when ± is plus. Add 2 to 10.
x=1
Divide 12 by 12.
x=-\frac{8}{12}
Now solve the equation x=\frac{2±10}{12} when ± is minus. Subtract 10 from 2.
x=-\frac{2}{3}
Reduce the fraction \frac{-8}{12} to lowest terms by extracting and canceling out 4.
x=1 x=-\frac{2}{3}
The equation is now solved.
6x^{2}-2x-4=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.
6x^{2}-2x-4-\left(-4\right)=-\left(-4\right)
Add 4 to both sides of the equation.
6x^{2}-2x=-\left(-4\right)
Subtracting -4 from itself leaves 0.
6x^{2}-2x=4
Subtract -4 from 0.
\frac{6x^{2}-2x}{6}=\frac{4}{6}
Divide both sides by 6.
x^{2}+\left(-\frac{2}{6}\right)x=\frac{4}{6}
Dividing by 6 undoes the multiplication by 6.
x^{2}-\frac{1}{3}x=\frac{4}{6}
Reduce the fraction \frac{-2}{6} to lowest terms by extracting and canceling out 2.
x^{2}-\frac{1}{3}x=\frac{2}{3}
Reduce the fraction \frac{4}{6} to lowest terms by extracting and canceling out 2.
x^{2}-\frac{1}{3}x+\left(-\frac{1}{6}\right)^{2}=\frac{2}{3}+\left(-\frac{1}{6}\right)^{2}
Divide -\frac{1}{3}, the coefficient of the x term, by 2 to get -\frac{1}{6}. Then add the square of -\frac{1}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{1}{3}x+\frac{1}{36}=\frac{2}{3}+\frac{1}{36}
Square -\frac{1}{6} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{1}{3}x+\frac{1}{36}=\frac{25}{36}
Add \frac{2}{3} to \frac{1}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{1}{6}\right)^{2}=\frac{25}{36}
Factor x^{2}-\frac{1}{3}x+\frac{1}{36}. 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}{6}\right)^{2}}=\sqrt{\frac{25}{36}}
Take the square root of both sides of the equation.
x-\frac{1}{6}=\frac{5}{6} x-\frac{1}{6}=-\frac{5}{6}
Simplify.
x=1 x=-\frac{2}{3}
Add \frac{1}{6} to both sides of the equation.
x ^ 2 -\frac{1}{3}x -\frac{2}{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 6
r + s = \frac{1}{3} rs = -\frac{2}{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}{6} - u s = \frac{1}{6} + u
Two numbers r and s sum up to \frac{1}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{1}{3} = \frac{1}{6}. 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}{6} - u) (\frac{1}{6} + u) = -\frac{2}{3}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{2}{3}
\frac{1}{36} - u^2 = -\frac{2}{3}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{2}{3}-\frac{1}{36} = -\frac{25}{36}
Simplify the expression by subtracting \frac{1}{36} on both sides
u^2 = \frac{25}{36} u = \pm\sqrt{\frac{25}{36}} = \pm \frac{5}{6}
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}{6} - \frac{5}{6} = -0.667 s = \frac{1}{6} + \frac{5}{6} = 1
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.