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4x^{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 4\times 4}}{2\times 4}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 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 4\times 4}}{2\times 4}
Square -2.
x=\frac{-\left(-2\right)±\sqrt{4-16\times 4}}{2\times 4}
Multiply -4 times 4.
x=\frac{-\left(-2\right)±\sqrt{4-64}}{2\times 4}
Multiply -16 times 4.
x=\frac{-\left(-2\right)±\sqrt{-60}}{2\times 4}
Add 4 to -64.
x=\frac{-\left(-2\right)±2\sqrt{15}i}{2\times 4}
Take the square root of -60.
x=\frac{2±2\sqrt{15}i}{2\times 4}
The opposite of -2 is 2.
x=\frac{2±2\sqrt{15}i}{8}
Multiply 2 times 4.
x=\frac{2+2\sqrt{15}i}{8}
Now solve the equation x=\frac{2±2\sqrt{15}i}{8} when ± is plus. Add 2 to 2i\sqrt{15}.
x=\frac{1+\sqrt{15}i}{4}
Divide 2+2i\sqrt{15} by 8.
x=\frac{-2\sqrt{15}i+2}{8}
Now solve the equation x=\frac{2±2\sqrt{15}i}{8} when ± is minus. Subtract 2i\sqrt{15} from 2.
x=\frac{-\sqrt{15}i+1}{4}
Divide 2-2i\sqrt{15} by 8.
x=\frac{1+\sqrt{15}i}{4} x=\frac{-\sqrt{15}i+1}{4}
The equation is now solved.
4x^{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.
4x^{2}-2x+4-4=-4
Subtract 4 from both sides of the equation.
4x^{2}-2x=-4
Subtracting 4 from itself leaves 0.
\frac{4x^{2}-2x}{4}=-\frac{4}{4}
Divide both sides by 4.
x^{2}+\left(-\frac{2}{4}\right)x=-\frac{4}{4}
Dividing by 4 undoes the multiplication by 4.
x^{2}-\frac{1}{2}x=-\frac{4}{4}
Reduce the fraction \frac{-2}{4} to lowest terms by extracting and canceling out 2.
x^{2}-\frac{1}{2}x=-1
Divide -4 by 4.
x^{2}-\frac{1}{2}x+\left(-\frac{1}{4}\right)^{2}=-1+\left(-\frac{1}{4}\right)^{2}
Divide -\frac{1}{2}, the coefficient of the x term, by 2 to get -\frac{1}{4}. Then add the square of -\frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{1}{2}x+\frac{1}{16}=-1+\frac{1}{16}
Square -\frac{1}{4} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{1}{2}x+\frac{1}{16}=-\frac{15}{16}
Add -1 to \frac{1}{16}.
\left(x-\frac{1}{4}\right)^{2}=-\frac{15}{16}
Factor x^{2}-\frac{1}{2}x+\frac{1}{16}. 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}{4}\right)^{2}}=\sqrt{-\frac{15}{16}}
Take the square root of both sides of the equation.
x-\frac{1}{4}=\frac{\sqrt{15}i}{4} x-\frac{1}{4}=-\frac{\sqrt{15}i}{4}
Simplify.
x=\frac{1+\sqrt{15}i}{4} x=\frac{-\sqrt{15}i+1}{4}
Add \frac{1}{4} to both sides of the equation.
x ^ 2 -\frac{1}{2}x +1 = 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 4
r + s = \frac{1}{2} rs = 1
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}{4} - u s = \frac{1}{4} + u
Two numbers r and s sum up to \frac{1}{2} exactly when the average of the two numbers is \frac{1}{2}*\frac{1}{2} = \frac{1}{4}. 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}{4} - u) (\frac{1}{4} + u) = 1
To solve for unknown quantity u, substitute these in the product equation rs = 1
\frac{1}{16} - u^2 = 1
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 1-\frac{1}{16} = \frac{15}{16}
Simplify the expression by subtracting \frac{1}{16} on both sides
u^2 = -\frac{15}{16} u = \pm\sqrt{-\frac{15}{16}} = \pm \frac{\sqrt{15}}{4}i
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}{4} - \frac{\sqrt{15}}{4}i = 0.250 - 0.968i s = \frac{1}{4} + \frac{\sqrt{15}}{4}i = 0.250 + 0.968i
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.