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-4x^{2}+8x-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{-8±\sqrt{8^{2}-4\left(-4\right)\left(-1\right)}}{2\left(-4\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, 8 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-8±\sqrt{64-4\left(-4\right)\left(-1\right)}}{2\left(-4\right)}
Square 8.
x=\frac{-8±\sqrt{64+16\left(-1\right)}}{2\left(-4\right)}
Multiply -4 times -4.
x=\frac{-8±\sqrt{64-16}}{2\left(-4\right)}
Multiply 16 times -1.
x=\frac{-8±\sqrt{48}}{2\left(-4\right)}
Add 64 to -16.
x=\frac{-8±4\sqrt{3}}{2\left(-4\right)}
Take the square root of 48.
x=\frac{-8±4\sqrt{3}}{-8}
Multiply 2 times -4.
x=\frac{4\sqrt{3}-8}{-8}
Now solve the equation x=\frac{-8±4\sqrt{3}}{-8} when ± is plus. Add -8 to 4\sqrt{3}.
x=-\frac{\sqrt{3}}{2}+1
Divide -8+4\sqrt{3} by -8.
x=\frac{-4\sqrt{3}-8}{-8}
Now solve the equation x=\frac{-8±4\sqrt{3}}{-8} when ± is minus. Subtract 4\sqrt{3} from -8.
x=\frac{\sqrt{3}}{2}+1
Divide -8-4\sqrt{3} by -8.
x=-\frac{\sqrt{3}}{2}+1 x=\frac{\sqrt{3}}{2}+1
The equation is now solved.
-4x^{2}+8x-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.
-4x^{2}+8x-1-\left(-1\right)=-\left(-1\right)
Add 1 to both sides of the equation.
-4x^{2}+8x=-\left(-1\right)
Subtracting -1 from itself leaves 0.
-4x^{2}+8x=1
Subtract -1 from 0.
\frac{-4x^{2}+8x}{-4}=\frac{1}{-4}
Divide both sides by -4.
x^{2}+\frac{8}{-4}x=\frac{1}{-4}
Dividing by -4 undoes the multiplication by -4.
x^{2}-2x=\frac{1}{-4}
Divide 8 by -4.
x^{2}-2x=-\frac{1}{4}
Divide 1 by -4.
x^{2}-2x+1=-\frac{1}{4}+1
Divide -2, the coefficient of the x term, by 2 to get -1. Then add the square of -1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-2x+1=\frac{3}{4}
Add -\frac{1}{4} to 1.
\left(x-1\right)^{2}=\frac{3}{4}
Factor x^{2}-2x+1. 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-1\right)^{2}}=\sqrt{\frac{3}{4}}
Take the square root of both sides of the equation.
x-1=\frac{\sqrt{3}}{2} x-1=-\frac{\sqrt{3}}{2}
Simplify.
x=\frac{\sqrt{3}}{2}+1 x=-\frac{\sqrt{3}}{2}+1
Add 1 to both sides of the equation.
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{3}{4}
Simplify the expression by subtracting 1 on both sides
u^2 = \frac{3}{4} u = \pm\sqrt{\frac{3}{4}} = \pm \frac{\sqrt{3}}{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{3}}{2} = 0.134 s = 1 + \frac{\sqrt{3}}{2} = 1.866
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.