Solve for h
h = \frac{\sqrt{5} + 1}{2} \approx 1.618033989
h=\frac{1-\sqrt{5}}{2}\approx -0.618033989
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-h^{2}+h+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{-1±\sqrt{1^{2}-4\left(-1\right)}}{2\left(-1\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 1 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
h=\frac{-1±\sqrt{1-4\left(-1\right)}}{2\left(-1\right)}
Square 1.
h=\frac{-1±\sqrt{1+4}}{2\left(-1\right)}
Multiply -4 times -1.
h=\frac{-1±\sqrt{5}}{2\left(-1\right)}
Add 1 to 4.
h=\frac{-1±\sqrt{5}}{-2}
Multiply 2 times -1.
h=\frac{\sqrt{5}-1}{-2}
Now solve the equation h=\frac{-1±\sqrt{5}}{-2} when ± is plus. Add -1 to \sqrt{5}.
h=\frac{1-\sqrt{5}}{2}
Divide -1+\sqrt{5} by -2.
h=\frac{-\sqrt{5}-1}{-2}
Now solve the equation h=\frac{-1±\sqrt{5}}{-2} when ± is minus. Subtract \sqrt{5} from -1.
h=\frac{\sqrt{5}+1}{2}
Divide -1-\sqrt{5} by -2.
h=\frac{1-\sqrt{5}}{2} h=\frac{\sqrt{5}+1}{2}
The equation is now solved.
-h^{2}+h+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.
-h^{2}+h+1-1=-1
Subtract 1 from both sides of the equation.
-h^{2}+h=-1
Subtracting 1 from itself leaves 0.
\frac{-h^{2}+h}{-1}=-\frac{1}{-1}
Divide both sides by -1.
h^{2}+\frac{1}{-1}h=-\frac{1}{-1}
Dividing by -1 undoes the multiplication by -1.
h^{2}-h=-\frac{1}{-1}
Divide 1 by -1.
h^{2}-h=1
Divide -1 by -1.
h^{2}-h+\left(-\frac{1}{2}\right)^{2}=1+\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}=1+\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{5}{4}
Add 1 to \frac{1}{4}.
\left(h-\frac{1}{2}\right)^{2}=\frac{5}{4}
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{5}{4}}
Take the square root of both sides of the equation.
h-\frac{1}{2}=\frac{\sqrt{5}}{2} h-\frac{1}{2}=-\frac{\sqrt{5}}{2}
Simplify.
h=\frac{\sqrt{5}+1}{2} h=\frac{1-\sqrt{5}}{2}
Add \frac{1}{2} to both sides of the equation.
x ^ 2 -1x -1 = 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 = -1
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) = -1
To solve for unknown quantity u, substitute these in the product equation rs = -1
\frac{1}{4} - u^2 = -1
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -1-\frac{1}{4} = -\frac{5}{4}
Simplify the expression by subtracting \frac{1}{4} on both sides
u^2 = \frac{5}{4} u = \pm\sqrt{\frac{5}{4}} = \pm \frac{\sqrt{5}}{2}
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{\sqrt{5}}{2} = -0.618 s = \frac{1}{2} + \frac{\sqrt{5}}{2} = 1.618
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
Arithmetic
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Matrix
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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
\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x
Limits
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