Solve for x
x=\frac{1}{4}=0.25
x=\frac{1}{2}=0.5
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a+b=6 ab=-8\left(-1\right)=8
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -8x^{2}+ax+bx-1. To find a and b, set up a system to be solved.
1,8 2,4
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 8.
1+8=9 2+4=6
Calculate the sum for each pair.
a=4 b=2
The solution is the pair that gives sum 6.
\left(-8x^{2}+4x\right)+\left(2x-1\right)
Rewrite -8x^{2}+6x-1 as \left(-8x^{2}+4x\right)+\left(2x-1\right).
-4x\left(2x-1\right)+2x-1
Factor out -4x in -8x^{2}+4x.
\left(2x-1\right)\left(-4x+1\right)
Factor out common term 2x-1 by using distributive property.
x=\frac{1}{2} x=\frac{1}{4}
To find equation solutions, solve 2x-1=0 and -4x+1=0.
-8x^{2}+6x-1=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{-6±\sqrt{6^{2}-4\left(-8\right)\left(-1\right)}}{2\left(-8\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -8 for a, 6 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-6±\sqrt{36-4\left(-8\right)\left(-1\right)}}{2\left(-8\right)}
Square 6.
x=\frac{-6±\sqrt{36+32\left(-1\right)}}{2\left(-8\right)}
Multiply -4 times -8.
x=\frac{-6±\sqrt{36-32}}{2\left(-8\right)}
Multiply 32 times -1.
x=\frac{-6±\sqrt{4}}{2\left(-8\right)}
Add 36 to -32.
x=\frac{-6±2}{2\left(-8\right)}
Take the square root of 4.
x=\frac{-6±2}{-16}
Multiply 2 times -8.
x=-\frac{4}{-16}
Now solve the equation x=\frac{-6±2}{-16} when ± is plus. Add -6 to 2.
x=\frac{1}{4}
Reduce the fraction \frac{-4}{-16} to lowest terms by extracting and canceling out 4.
x=-\frac{8}{-16}
Now solve the equation x=\frac{-6±2}{-16} when ± is minus. Subtract 2 from -6.
x=\frac{1}{2}
Reduce the fraction \frac{-8}{-16} to lowest terms by extracting and canceling out 8.
x=\frac{1}{4} x=\frac{1}{2}
The equation is now solved.
-8x^{2}+6x-1=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.
-8x^{2}+6x-1-\left(-1\right)=-\left(-1\right)
Add 1 to both sides of the equation.
-8x^{2}+6x=-\left(-1\right)
Subtracting -1 from itself leaves 0.
-8x^{2}+6x=1
Subtract -1 from 0.
\frac{-8x^{2}+6x}{-8}=\frac{1}{-8}
Divide both sides by -8.
x^{2}+\frac{6}{-8}x=\frac{1}{-8}
Dividing by -8 undoes the multiplication by -8.
x^{2}-\frac{3}{4}x=\frac{1}{-8}
Reduce the fraction \frac{6}{-8} to lowest terms by extracting and canceling out 2.
x^{2}-\frac{3}{4}x=-\frac{1}{8}
Divide 1 by -8.
x^{2}-\frac{3}{4}x+\left(-\frac{3}{8}\right)^{2}=-\frac{1}{8}+\left(-\frac{3}{8}\right)^{2}
Divide -\frac{3}{4}, the coefficient of the x term, by 2 to get -\frac{3}{8}. Then add the square of -\frac{3}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{3}{4}x+\frac{9}{64}=-\frac{1}{8}+\frac{9}{64}
Square -\frac{3}{8} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{3}{4}x+\frac{9}{64}=\frac{1}{64}
Add -\frac{1}{8} to \frac{9}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{3}{8}\right)^{2}=\frac{1}{64}
Factor x^{2}-\frac{3}{4}x+\frac{9}{64}. 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{3}{8}\right)^{2}}=\sqrt{\frac{1}{64}}
Take the square root of both sides of the equation.
x-\frac{3}{8}=\frac{1}{8} x-\frac{3}{8}=-\frac{1}{8}
Simplify.
x=\frac{1}{2} x=\frac{1}{4}
Add \frac{3}{8} to both sides of the equation.
x ^ 2 -\frac{3}{4}x +\frac{1}{8} = 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.
r + s = \frac{3}{4} rs = \frac{1}{8}
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{3}{8} - u s = \frac{3}{8} + u
Two numbers r and s sum up to \frac{3}{4} exactly when the average of the two numbers is \frac{1}{2}*\frac{3}{4} = \frac{3}{8}. 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{3}{8} - u) (\frac{3}{8} + u) = \frac{1}{8}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{1}{8}
\frac{9}{64} - u^2 = \frac{1}{8}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{1}{8}-\frac{9}{64} = -\frac{1}{64}
Simplify the expression by subtracting \frac{9}{64} on both sides
u^2 = \frac{1}{64} u = \pm\sqrt{\frac{1}{64}} = \pm \frac{1}{8}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{3}{8} - \frac{1}{8} = 0.250 s = \frac{3}{8} + \frac{1}{8} = 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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