Solve for x
x=\frac{3\sqrt{5}}{10}+\frac{1}{2}\approx 1.170820393
x=-\frac{3\sqrt{5}}{10}+\frac{1}{2}\approx -0.170820393
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5x^{2}-5x-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{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 5\left(-1\right)}}{2\times 5}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -5 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-5\right)±\sqrt{25-4\times 5\left(-1\right)}}{2\times 5}
Square -5.
x=\frac{-\left(-5\right)±\sqrt{25-20\left(-1\right)}}{2\times 5}
Multiply -4 times 5.
x=\frac{-\left(-5\right)±\sqrt{25+20}}{2\times 5}
Multiply -20 times -1.
x=\frac{-\left(-5\right)±\sqrt{45}}{2\times 5}
Add 25 to 20.
x=\frac{-\left(-5\right)±3\sqrt{5}}{2\times 5}
Take the square root of 45.
x=\frac{5±3\sqrt{5}}{2\times 5}
The opposite of -5 is 5.
x=\frac{5±3\sqrt{5}}{10}
Multiply 2 times 5.
x=\frac{3\sqrt{5}+5}{10}
Now solve the equation x=\frac{5±3\sqrt{5}}{10} when ± is plus. Add 5 to 3\sqrt{5}.
x=\frac{3\sqrt{5}}{10}+\frac{1}{2}
Divide 5+3\sqrt{5} by 10.
x=\frac{5-3\sqrt{5}}{10}
Now solve the equation x=\frac{5±3\sqrt{5}}{10} when ± is minus. Subtract 3\sqrt{5} from 5.
x=-\frac{3\sqrt{5}}{10}+\frac{1}{2}
Divide 5-3\sqrt{5} by 10.
x=\frac{3\sqrt{5}}{10}+\frac{1}{2} x=-\frac{3\sqrt{5}}{10}+\frac{1}{2}
The equation is now solved.
5x^{2}-5x-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.
5x^{2}-5x-1-\left(-1\right)=-\left(-1\right)
Add 1 to both sides of the equation.
5x^{2}-5x=-\left(-1\right)
Subtracting -1 from itself leaves 0.
5x^{2}-5x=1
Subtract -1 from 0.
\frac{5x^{2}-5x}{5}=\frac{1}{5}
Divide both sides by 5.
x^{2}+\left(-\frac{5}{5}\right)x=\frac{1}{5}
Dividing by 5 undoes the multiplication by 5.
x^{2}-x=\frac{1}{5}
Divide -5 by 5.
x^{2}-x+\left(-\frac{1}{2}\right)^{2}=\frac{1}{5}+\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}{5}+\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{9}{20}
Add \frac{1}{5} 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{9}{20}
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{9}{20}}
Take the square root of both sides of the equation.
x-\frac{1}{2}=\frac{3\sqrt{5}}{10} x-\frac{1}{2}=-\frac{3\sqrt{5}}{10}
Simplify.
x=\frac{3\sqrt{5}}{10}+\frac{1}{2} x=-\frac{3\sqrt{5}}{10}+\frac{1}{2}
Add \frac{1}{2} to both sides of the equation.
x ^ 2 -1x -\frac{1}{5} = 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 5
r + s = 1 rs = -\frac{1}{5}
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}{5}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{1}{5}
\frac{1}{4} - u^2 = -\frac{1}{5}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{1}{5}-\frac{1}{4} = -\frac{9}{20}
Simplify the expression by subtracting \frac{1}{4} on both sides
u^2 = \frac{9}{20} u = \pm\sqrt{\frac{9}{20}} = \pm \frac{3}{\sqrt{20}}
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{3}{\sqrt{20}} = -0.171 s = \frac{1}{2} + \frac{3}{\sqrt{20}} = 1.171
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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{ 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}