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2x^{2}+2x+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{-2±\sqrt{2^{2}-4\times 2}}{2\times 2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 2 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-2±\sqrt{4-4\times 2}}{2\times 2}
Square 2.
x=\frac{-2±\sqrt{4-8}}{2\times 2}
Multiply -4 times 2.
x=\frac{-2±\sqrt{-4}}{2\times 2}
Add 4 to -8.
x=\frac{-2±2i}{2\times 2}
Take the square root of -4.
x=\frac{-2±2i}{4}
Multiply 2 times 2.
x=\frac{-2+2i}{4}
Now solve the equation x=\frac{-2±2i}{4} when ± is plus. Add -2 to 2i.
x=-\frac{1}{2}+\frac{1}{2}i
Divide -2+2i by 4.
x=\frac{-2-2i}{4}
Now solve the equation x=\frac{-2±2i}{4} when ± is minus. Subtract 2i from -2.
x=-\frac{1}{2}-\frac{1}{2}i
Divide -2-2i by 4.
x=-\frac{1}{2}+\frac{1}{2}i x=-\frac{1}{2}-\frac{1}{2}i
The equation is now solved.
2x^{2}+2x+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.
2x^{2}+2x+1-1=-1
Subtract 1 from both sides of the equation.
2x^{2}+2x=-1
Subtracting 1 from itself leaves 0.
\frac{2x^{2}+2x}{2}=-\frac{1}{2}
Divide both sides by 2.
x^{2}+\frac{2}{2}x=-\frac{1}{2}
Dividing by 2 undoes the multiplication by 2.
x^{2}+x=-\frac{1}{2}
Divide 2 by 2.
x^{2}+x+\left(\frac{1}{2}\right)^{2}=-\frac{1}{2}+\left(\frac{1}{2}\right)^{2}
Divide 1, the coefficient of the x term, by 2 to get \frac{1}{2}. Then add the square of \frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+x+\frac{1}{4}=-\frac{1}{2}+\frac{1}{4}
Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.
x^{2}+x+\frac{1}{4}=-\frac{1}{4}
Add -\frac{1}{2} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x+\frac{1}{2}\right)^{2}=-\frac{1}{4}
Factor x^{2}+x+\frac{1}{4}. 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}{2}\right)^{2}}=\sqrt{-\frac{1}{4}}
Take the square root of both sides of the equation.
x+\frac{1}{2}=\frac{1}{2}i x+\frac{1}{2}=-\frac{1}{2}i
Simplify.
x=-\frac{1}{2}+\frac{1}{2}i x=-\frac{1}{2}-\frac{1}{2}i
Subtract \frac{1}{2} from both sides of the equation.
x ^ 2 +1x +\frac{1}{2} = 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 2
r + s = -1 rs = \frac{1}{2}
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}{2} - u s = -\frac{1}{2} + u
Two numbers r and s sum up to -1 exactly when the average of the two numbers is \frac{1}{2}*-1 = -\frac{1}{2}. 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}{2} - u) (-\frac{1}{2} + u) = \frac{1}{2}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{1}{2}
\frac{1}{4} - u^2 = \frac{1}{2}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{1}{2}-\frac{1}{4} = \frac{1}{4}
Simplify the expression by subtracting \frac{1}{4} on both sides
u^2 = -\frac{1}{4} u = \pm\sqrt{-\frac{1}{4}} = \pm \frac{1}{2}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}{2} - \frac{1}{2}i = -0.500 - 0.500i s = -\frac{1}{2} + \frac{1}{2}i = -0.500 + 0.500i
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.