Solve for x
x=\frac{\sqrt{6}}{3}+1\approx 1.816496581
x=-\frac{\sqrt{6}}{3}+1\approx 0.183503419
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-9x^{2}+18x-3=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{-18±\sqrt{18^{2}-4\left(-9\right)\left(-3\right)}}{2\left(-9\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -9 for a, 18 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-18±\sqrt{324-4\left(-9\right)\left(-3\right)}}{2\left(-9\right)}
Square 18.
x=\frac{-18±\sqrt{324+36\left(-3\right)}}{2\left(-9\right)}
Multiply -4 times -9.
x=\frac{-18±\sqrt{324-108}}{2\left(-9\right)}
Multiply 36 times -3.
x=\frac{-18±\sqrt{216}}{2\left(-9\right)}
Add 324 to -108.
x=\frac{-18±6\sqrt{6}}{2\left(-9\right)}
Take the square root of 216.
x=\frac{-18±6\sqrt{6}}{-18}
Multiply 2 times -9.
x=\frac{6\sqrt{6}-18}{-18}
Now solve the equation x=\frac{-18±6\sqrt{6}}{-18} when ± is plus. Add -18 to 6\sqrt{6}.
x=-\frac{\sqrt{6}}{3}+1
Divide -18+6\sqrt{6} by -18.
x=\frac{-6\sqrt{6}-18}{-18}
Now solve the equation x=\frac{-18±6\sqrt{6}}{-18} when ± is minus. Subtract 6\sqrt{6} from -18.
x=\frac{\sqrt{6}}{3}+1
Divide -18-6\sqrt{6} by -18.
x=-\frac{\sqrt{6}}{3}+1 x=\frac{\sqrt{6}}{3}+1
The equation is now solved.
-9x^{2}+18x-3=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.
-9x^{2}+18x-3-\left(-3\right)=-\left(-3\right)
Add 3 to both sides of the equation.
-9x^{2}+18x=-\left(-3\right)
Subtracting -3 from itself leaves 0.
-9x^{2}+18x=3
Subtract -3 from 0.
\frac{-9x^{2}+18x}{-9}=\frac{3}{-9}
Divide both sides by -9.
x^{2}+\frac{18}{-9}x=\frac{3}{-9}
Dividing by -9 undoes the multiplication by -9.
x^{2}-2x=\frac{3}{-9}
Divide 18 by -9.
x^{2}-2x=-\frac{1}{3}
Reduce the fraction \frac{3}{-9} to lowest terms by extracting and canceling out 3.
x^{2}-2x+1=-\frac{1}{3}+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=\frac{2}{3}
Add -\frac{1}{3} to 1.
\left(x-1\right)^{2}=\frac{2}{3}
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{\frac{2}{3}}
Take the square root of both sides of the equation.
x-1=\frac{\sqrt{6}}{3} x-1=-\frac{\sqrt{6}}{3}
Simplify.
x=\frac{\sqrt{6}}{3}+1 x=-\frac{\sqrt{6}}{3}+1
Add 1 to both sides of the equation.
x ^ 2 -2x +\frac{1}{3} = 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 = \frac{1}{3}
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) = \frac{1}{3}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{1}{3}
1 - u^2 = \frac{1}{3}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{1}{3}-1 = -\frac{2}{3}
Simplify the expression by subtracting 1 on both sides
u^2 = \frac{2}{3} u = \pm\sqrt{\frac{2}{3}} = \pm \frac{\sqrt{2}}{\sqrt{3}}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =1 - \frac{\sqrt{2}}{\sqrt{3}} = 0.184 s = 1 + \frac{\sqrt{2}}{\sqrt{3}} = 1.816
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
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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
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