Solve for x (complex solution)
x=1+i
x=1-i
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x^{2}-2x+2=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 2}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -2 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-2\right)±\sqrt{4-4\times 2}}{2}
Square -2.
x=\frac{-\left(-2\right)±\sqrt{4-8}}{2}
Multiply -4 times 2.
x=\frac{-\left(-2\right)±\sqrt{-4}}{2}
Add 4 to -8.
x=\frac{-\left(-2\right)±2i}{2}
Take the square root of -4.
x=\frac{2±2i}{2}
The opposite of -2 is 2.
x=\frac{2+2i}{2}
Now solve the equation x=\frac{2±2i}{2} when ± is plus. Add 2 to 2i.
x=1+i
Divide 2+2i by 2.
x=\frac{2-2i}{2}
Now solve the equation x=\frac{2±2i}{2} when ± is minus. Subtract 2i from 2.
x=1-i
Divide 2-2i by 2.
x=1+i x=1-i
The equation is now solved.
x^{2}-2x+2=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.
x^{2}-2x+2-2=-2
Subtract 2 from both sides of the equation.
x^{2}-2x=-2
Subtracting 2 from itself leaves 0.
x^{2}-2x+1=-2+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=-1
Add -2 to 1.
\left(x-1\right)^{2}=-1
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{-1}
Take the square root of both sides of the equation.
x-1=i x-1=-i
Simplify.
x=1+i x=1-i
Add 1 to both sides of the equation.
x ^ 2 -2x +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.
r + s = 2 rs = 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 = 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) = 2
To solve for unknown quantity u, substitute these in the product equation rs = 2
1 - u^2 = 2
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 2-1 = 1
Simplify the expression by subtracting 1 on both sides
u^2 = -1 u = \pm\sqrt{-1} = \pm i
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =1 - i s = 1 + i
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
Examples
Quadratic equation
{ x } ^ { 2 } - 4 x - 5 = 0
Trigonometry
4 \sin \theta \cos \theta = 2 \sin \theta
Linear equation
y = 3x + 4
Arithmetic
699 * 533
Matrix
\left[ \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \right] \left[ \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { -1 } & { 1 } & { 5 } \end{array} \right]
Simultaneous equation
\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.
Differentiation
\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }
Integration
\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x
Limits
\lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}