Solve for x
x=-\frac{2}{3}\approx -0.666666667
x=-\frac{1}{2}=-0.5
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a+b=7 ab=6\times 2=12
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 6x^{2}+ax+bx+2. To find a and b, set up a system to be solved.
1,12 2,6 3,4
Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 12.
1+12=13 2+6=8 3+4=7
Calculate the sum for each pair.
a=3 b=4
The solution is the pair that gives sum 7.
\left(6x^{2}+3x\right)+\left(4x+2\right)
Rewrite 6x^{2}+7x+2 as \left(6x^{2}+3x\right)+\left(4x+2\right).
3x\left(2x+1\right)+2\left(2x+1\right)
Factor out 3x in the first and 2 in the second group.
\left(2x+1\right)\left(3x+2\right)
Factor out common term 2x+1 by using distributive property.
x=-\frac{1}{2} x=-\frac{2}{3}
To find equation solutions, solve 2x+1=0 and 3x+2=0.
6x^{2}+7x+2=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{-7±\sqrt{7^{2}-4\times 6\times 2}}{2\times 6}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, 7 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-7±\sqrt{49-4\times 6\times 2}}{2\times 6}
Square 7.
x=\frac{-7±\sqrt{49-24\times 2}}{2\times 6}
Multiply -4 times 6.
x=\frac{-7±\sqrt{49-48}}{2\times 6}
Multiply -24 times 2.
x=\frac{-7±\sqrt{1}}{2\times 6}
Add 49 to -48.
x=\frac{-7±1}{2\times 6}
Take the square root of 1.
x=\frac{-7±1}{12}
Multiply 2 times 6.
x=-\frac{6}{12}
Now solve the equation x=\frac{-7±1}{12} when ± is plus. Add -7 to 1.
x=-\frac{1}{2}
Reduce the fraction \frac{-6}{12} to lowest terms by extracting and canceling out 6.
x=-\frac{8}{12}
Now solve the equation x=\frac{-7±1}{12} when ± is minus. Subtract 1 from -7.
x=-\frac{2}{3}
Reduce the fraction \frac{-8}{12} to lowest terms by extracting and canceling out 4.
x=-\frac{1}{2} x=-\frac{2}{3}
The equation is now solved.
6x^{2}+7x+2=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}+7x+2-2=-2
Subtract 2 from both sides of the equation.
6x^{2}+7x=-2
Subtracting 2 from itself leaves 0.
\frac{6x^{2}+7x}{6}=-\frac{2}{6}
Divide both sides by 6.
x^{2}+\frac{7}{6}x=-\frac{2}{6}
Dividing by 6 undoes the multiplication by 6.
x^{2}+\frac{7}{6}x=-\frac{1}{3}
Reduce the fraction \frac{-2}{6} to lowest terms by extracting and canceling out 2.
x^{2}+\frac{7}{6}x+\left(\frac{7}{12}\right)^{2}=-\frac{1}{3}+\left(\frac{7}{12}\right)^{2}
Divide \frac{7}{6}, the coefficient of the x term, by 2 to get \frac{7}{12}. Then add the square of \frac{7}{12} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+\frac{7}{6}x+\frac{49}{144}=-\frac{1}{3}+\frac{49}{144}
Square \frac{7}{12} by squaring both the numerator and the denominator of the fraction.
x^{2}+\frac{7}{6}x+\frac{49}{144}=\frac{1}{144}
Add -\frac{1}{3} to \frac{49}{144} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x+\frac{7}{12}\right)^{2}=\frac{1}{144}
Factor x^{2}+\frac{7}{6}x+\frac{49}{144}. 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{7}{12}\right)^{2}}=\sqrt{\frac{1}{144}}
Take the square root of both sides of the equation.
x+\frac{7}{12}=\frac{1}{12} x+\frac{7}{12}=-\frac{1}{12}
Simplify.
x=-\frac{1}{2} x=-\frac{2}{3}
Subtract \frac{7}{12} from both sides of the equation.
x ^ 2 +\frac{7}{6}x +\frac{1}{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{7}{6} rs = \frac{1}{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{7}{12} - u s = -\frac{7}{12} + u
Two numbers r and s sum up to -\frac{7}{6} exactly when the average of the two numbers is \frac{1}{2}*-\frac{7}{6} = -\frac{7}{12}. 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{7}{12} - u) (-\frac{7}{12} + u) = \frac{1}{3}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{1}{3}
\frac{49}{144} - u^2 = \frac{1}{3}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{1}{3}-\frac{49}{144} = -\frac{1}{144}
Simplify the expression by subtracting \frac{49}{144} on both sides
u^2 = \frac{1}{144} u = \pm\sqrt{\frac{1}{144}} = \pm \frac{1}{12}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-\frac{7}{12} - \frac{1}{12} = -0.667 s = -\frac{7}{12} + \frac{1}{12} = -0.500
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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