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-4x^{2}-8x+1=0
Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.
x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\left(-4\right)}}{2\left(-4\right)}
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(-8\right)±\sqrt{64-4\left(-4\right)}}{2\left(-4\right)}
Square -8.
x=\frac{-\left(-8\right)±\sqrt{64+16}}{2\left(-4\right)}
Multiply -4 times -4.
x=\frac{-\left(-8\right)±\sqrt{80}}{2\left(-4\right)}
Add 64 to 16.
x=\frac{-\left(-8\right)±4\sqrt{5}}{2\left(-4\right)}
Take the square root of 80.
x=\frac{8±4\sqrt{5}}{2\left(-4\right)}
The opposite of -8 is 8.
x=\frac{8±4\sqrt{5}}{-8}
Multiply 2 times -4.
x=\frac{4\sqrt{5}+8}{-8}
Now solve the equation x=\frac{8±4\sqrt{5}}{-8} when ± is plus. Add 8 to 4\sqrt{5}.
x=-\frac{\sqrt{5}}{2}-1
Divide 8+4\sqrt{5} by -8.
x=\frac{8-4\sqrt{5}}{-8}
Now solve the equation x=\frac{8±4\sqrt{5}}{-8} when ± is minus. Subtract 4\sqrt{5} from 8.
x=\frac{\sqrt{5}}{2}-1
Divide 8-4\sqrt{5} by -8.
-4x^{2}-8x+1=-4\left(x-\left(-\frac{\sqrt{5}}{2}-1\right)\right)\left(x-\left(\frac{\sqrt{5}}{2}-1\right)\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -1-\frac{\sqrt{5}}{2} for x_{1} and -1+\frac{\sqrt{5}}{2} for x_{2}.
x ^ 2 +2x -\frac{1}{4} = 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 = -2 rs = -\frac{1}{4}
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 = -1 - u s = -1 + u
Two numbers r and s sum up to -2 exactly when the average of the two numbers is \frac{1}{2}*-2 = -1. 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>
(-1 - u) (-1 + u) = -\frac{1}{4}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{1}{4}
1 - u^2 = -\frac{1}{4}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{1}{4}-1 = -\frac{5}{4}
Simplify the expression by subtracting 1 on both sides
u^2 = \frac{5}{4} u = \pm\sqrt{\frac{5}{4}} = \pm \frac{\sqrt{5}}{2}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-1 - \frac{\sqrt{5}}{2} = -2.118 s = -1 + \frac{\sqrt{5}}{2} = 0.118
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.