Solve for x
x=-\frac{1}{2}=-0.5
x=1
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2x^{2}-x-1=0
Divide both sides by 2.
a+b=-1 ab=2\left(-1\right)=-2
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx-1. To find a and b, set up a system to be solved.
a=-2 b=1
Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. The only such pair is the system solution.
\left(2x^{2}-2x\right)+\left(x-1\right)
Rewrite 2x^{2}-x-1 as \left(2x^{2}-2x\right)+\left(x-1\right).
2x\left(x-1\right)+x-1
Factor out 2x in 2x^{2}-2x.
\left(x-1\right)\left(2x+1\right)
Factor out common term x-1 by using distributive property.
x=1 x=-\frac{1}{2}
To find equation solutions, solve x-1=0 and 2x+1=0.
4x^{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 4\left(-2\right)}}{2\times 4}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 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 4\left(-2\right)}}{2\times 4}
Square -2.
x=\frac{-\left(-2\right)±\sqrt{4-16\left(-2\right)}}{2\times 4}
Multiply -4 times 4.
x=\frac{-\left(-2\right)±\sqrt{4+32}}{2\times 4}
Multiply -16 times -2.
x=\frac{-\left(-2\right)±\sqrt{36}}{2\times 4}
Add 4 to 32.
x=\frac{-\left(-2\right)±6}{2\times 4}
Take the square root of 36.
x=\frac{2±6}{2\times 4}
The opposite of -2 is 2.
x=\frac{2±6}{8}
Multiply 2 times 4.
x=\frac{8}{8}
Now solve the equation x=\frac{2±6}{8} when ± is plus. Add 2 to 6.
x=1
Divide 8 by 8.
x=-\frac{4}{8}
Now solve the equation x=\frac{2±6}{8} when ± is minus. Subtract 6 from 2.
x=-\frac{1}{2}
Reduce the fraction \frac{-4}{8} to lowest terms by extracting and canceling out 4.
x=1 x=-\frac{1}{2}
The equation is now solved.
4x^{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.
4x^{2}-2x-2-\left(-2\right)=-\left(-2\right)
Add 2 to both sides of the equation.
4x^{2}-2x=-\left(-2\right)
Subtracting -2 from itself leaves 0.
4x^{2}-2x=2
Subtract -2 from 0.
\frac{4x^{2}-2x}{4}=\frac{2}{4}
Divide both sides by 4.
x^{2}+\left(-\frac{2}{4}\right)x=\frac{2}{4}
Dividing by 4 undoes the multiplication by 4.
x^{2}-\frac{1}{2}x=\frac{2}{4}
Reduce the fraction \frac{-2}{4} to lowest terms by extracting and canceling out 2.
x^{2}-\frac{1}{2}x=\frac{1}{2}
Reduce the fraction \frac{2}{4} to lowest terms by extracting and canceling out 2.
x^{2}-\frac{1}{2}x+\left(-\frac{1}{4}\right)^{2}=\frac{1}{2}+\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}=\frac{1}{2}+\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{9}{16}
Add \frac{1}{2} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{1}{4}\right)^{2}=\frac{9}{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{9}{16}}
Take the square root of both sides of the equation.
x-\frac{1}{4}=\frac{3}{4} x-\frac{1}{4}=-\frac{3}{4}
Simplify.
x=1 x=-\frac{1}{2}
Add \frac{1}{4} to both sides of the equation.
x ^ 2 -\frac{1}{2}x -\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 4
r + s = \frac{1}{2} 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}{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) = -\frac{1}{2}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{1}{2}
\frac{1}{16} - u^2 = -\frac{1}{2}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{1}{2}-\frac{1}{16} = -\frac{9}{16}
Simplify the expression by subtracting \frac{1}{16} on both sides
u^2 = \frac{9}{16} u = \pm\sqrt{\frac{9}{16}} = \pm \frac{3}{4}
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{3}{4} = -0.500 s = \frac{1}{4} + \frac{3}{4} = 1
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
Examples
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Linear equation
y = 3x + 4
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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
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Limits
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