Solve for h
h=-\frac{\sqrt{3}i}{6}+\frac{1}{2}\approx 0.5-0.288675135i
h=\frac{\sqrt{3}i}{6}+\frac{1}{2}\approx 0.5+0.288675135i
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-3h^{2}+3h-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.
h=\frac{-3±\sqrt{3^{2}-4\left(-3\right)\left(-1\right)}}{2\left(-3\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 3 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
h=\frac{-3±\sqrt{9-4\left(-3\right)\left(-1\right)}}{2\left(-3\right)}
Square 3.
h=\frac{-3±\sqrt{9+12\left(-1\right)}}{2\left(-3\right)}
Multiply -4 times -3.
h=\frac{-3±\sqrt{9-12}}{2\left(-3\right)}
Multiply 12 times -1.
h=\frac{-3±\sqrt{-3}}{2\left(-3\right)}
Add 9 to -12.
h=\frac{-3±\sqrt{3}i}{2\left(-3\right)}
Take the square root of -3.
h=\frac{-3±\sqrt{3}i}{-6}
Multiply 2 times -3.
h=\frac{-3+\sqrt{3}i}{-6}
Now solve the equation h=\frac{-3±\sqrt{3}i}{-6} when ± is plus. Add -3 to i\sqrt{3}.
h=-\frac{\sqrt{3}i}{6}+\frac{1}{2}
Divide -3+i\sqrt{3} by -6.
h=\frac{-\sqrt{3}i-3}{-6}
Now solve the equation h=\frac{-3±\sqrt{3}i}{-6} when ± is minus. Subtract i\sqrt{3} from -3.
h=\frac{\sqrt{3}i}{6}+\frac{1}{2}
Divide -3-i\sqrt{3} by -6.
h=-\frac{\sqrt{3}i}{6}+\frac{1}{2} h=\frac{\sqrt{3}i}{6}+\frac{1}{2}
The equation is now solved.
-3h^{2}+3h-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.
-3h^{2}+3h-1-\left(-1\right)=-\left(-1\right)
Add 1 to both sides of the equation.
-3h^{2}+3h=-\left(-1\right)
Subtracting -1 from itself leaves 0.
-3h^{2}+3h=1
Subtract -1 from 0.
\frac{-3h^{2}+3h}{-3}=\frac{1}{-3}
Divide both sides by -3.
h^{2}+\frac{3}{-3}h=\frac{1}{-3}
Dividing by -3 undoes the multiplication by -3.
h^{2}-h=\frac{1}{-3}
Divide 3 by -3.
h^{2}-h=-\frac{1}{3}
Divide 1 by -3.
h^{2}-h+\left(-\frac{1}{2}\right)^{2}=-\frac{1}{3}+\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.
h^{2}-h+\frac{1}{4}=-\frac{1}{3}+\frac{1}{4}
Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.
h^{2}-h+\frac{1}{4}=-\frac{1}{12}
Add -\frac{1}{3} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(h-\frac{1}{2}\right)^{2}=-\frac{1}{12}
Factor h^{2}-h+\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(h-\frac{1}{2}\right)^{2}}=\sqrt{-\frac{1}{12}}
Take the square root of both sides of the equation.
h-\frac{1}{2}=\frac{\sqrt{3}i}{6} h-\frac{1}{2}=-\frac{\sqrt{3}i}{6}
Simplify.
h=\frac{\sqrt{3}i}{6}+\frac{1}{2} h=-\frac{\sqrt{3}i}{6}+\frac{1}{2}
Add \frac{1}{2} to both sides of the equation.
x ^ 2 -1x +\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 = 1 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 = \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}{3}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{1}{3}
\frac{1}{4} - u^2 = \frac{1}{3}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{1}{3}-\frac{1}{4} = \frac{1}{12}
Simplify the expression by subtracting \frac{1}{4} on both sides
u^2 = -\frac{1}{12} u = \pm\sqrt{-\frac{1}{12}} = \pm \frac{1}{\sqrt{12}}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}{\sqrt{12}}i = 0.500 - 0.289i s = \frac{1}{2} + \frac{1}{\sqrt{12}}i = 0.500 + 0.289i
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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