Solve for x
x = \frac{5 \sqrt{577} + 121}{12} \approx 20.092010125
x=\frac{121-5\sqrt{577}}{12}\approx 0.074656542
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-6x^{2}+121x-9=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{-121±\sqrt{121^{2}-4\left(-6\right)\left(-9\right)}}{2\left(-6\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -6 for a, 121 for b, and -9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-121±\sqrt{14641-4\left(-6\right)\left(-9\right)}}{2\left(-6\right)}
Square 121.
x=\frac{-121±\sqrt{14641+24\left(-9\right)}}{2\left(-6\right)}
Multiply -4 times -6.
x=\frac{-121±\sqrt{14641-216}}{2\left(-6\right)}
Multiply 24 times -9.
x=\frac{-121±\sqrt{14425}}{2\left(-6\right)}
Add 14641 to -216.
x=\frac{-121±5\sqrt{577}}{2\left(-6\right)}
Take the square root of 14425.
x=\frac{-121±5\sqrt{577}}{-12}
Multiply 2 times -6.
x=\frac{5\sqrt{577}-121}{-12}
Now solve the equation x=\frac{-121±5\sqrt{577}}{-12} when ± is plus. Add -121 to 5\sqrt{577}.
x=\frac{121-5\sqrt{577}}{12}
Divide -121+5\sqrt{577} by -12.
x=\frac{-5\sqrt{577}-121}{-12}
Now solve the equation x=\frac{-121±5\sqrt{577}}{-12} when ± is minus. Subtract 5\sqrt{577} from -121.
x=\frac{5\sqrt{577}+121}{12}
Divide -121-5\sqrt{577} by -12.
x=\frac{121-5\sqrt{577}}{12} x=\frac{5\sqrt{577}+121}{12}
The equation is now solved.
-6x^{2}+121x-9=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.
-6x^{2}+121x-9-\left(-9\right)=-\left(-9\right)
Add 9 to both sides of the equation.
-6x^{2}+121x=-\left(-9\right)
Subtracting -9 from itself leaves 0.
-6x^{2}+121x=9
Subtract -9 from 0.
\frac{-6x^{2}+121x}{-6}=\frac{9}{-6}
Divide both sides by -6.
x^{2}+\frac{121}{-6}x=\frac{9}{-6}
Dividing by -6 undoes the multiplication by -6.
x^{2}-\frac{121}{6}x=\frac{9}{-6}
Divide 121 by -6.
x^{2}-\frac{121}{6}x=-\frac{3}{2}
Reduce the fraction \frac{9}{-6} to lowest terms by extracting and canceling out 3.
x^{2}-\frac{121}{6}x+\left(-\frac{121}{12}\right)^{2}=-\frac{3}{2}+\left(-\frac{121}{12}\right)^{2}
Divide -\frac{121}{6}, the coefficient of the x term, by 2 to get -\frac{121}{12}. Then add the square of -\frac{121}{12} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{121}{6}x+\frac{14641}{144}=-\frac{3}{2}+\frac{14641}{144}
Square -\frac{121}{12} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{121}{6}x+\frac{14641}{144}=\frac{14425}{144}
Add -\frac{3}{2} to \frac{14641}{144} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{121}{12}\right)^{2}=\frac{14425}{144}
Factor x^{2}-\frac{121}{6}x+\frac{14641}{144}. 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{121}{12}\right)^{2}}=\sqrt{\frac{14425}{144}}
Take the square root of both sides of the equation.
x-\frac{121}{12}=\frac{5\sqrt{577}}{12} x-\frac{121}{12}=-\frac{5\sqrt{577}}{12}
Simplify.
x=\frac{5\sqrt{577}+121}{12} x=\frac{121-5\sqrt{577}}{12}
Add \frac{121}{12} to both sides of the equation.
x ^ 2 -\frac{121}{6}x +\frac{3}{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.
r + s = \frac{121}{6} rs = \frac{3}{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{121}{12} - u s = \frac{121}{12} + u
Two numbers r and s sum up to \frac{121}{6} exactly when the average of the two numbers is \frac{1}{2}*\frac{121}{6} = \frac{121}{12}. 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{121}{12} - u) (\frac{121}{12} + u) = \frac{3}{2}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{3}{2}
\frac{14641}{144} - u^2 = \frac{3}{2}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{3}{2}-\frac{14641}{144} = -\frac{14425}{144}
Simplify the expression by subtracting \frac{14641}{144} on both sides
u^2 = \frac{14425}{144} u = \pm\sqrt{\frac{14425}{144}} = \pm \frac{\sqrt{14425}}{12}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{121}{12} - \frac{\sqrt{14425}}{12} = 0.075 s = \frac{121}{12} + \frac{\sqrt{14425}}{12} = 20.092
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
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