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13a^{2}-12a-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.
a=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 13\left(-9\right)}}{2\times 13}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 13 for a, -12 for b, and -9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
a=\frac{-\left(-12\right)±\sqrt{144-4\times 13\left(-9\right)}}{2\times 13}
Square -12.
a=\frac{-\left(-12\right)±\sqrt{144-52\left(-9\right)}}{2\times 13}
Multiply -4 times 13.
a=\frac{-\left(-12\right)±\sqrt{144+468}}{2\times 13}
Multiply -52 times -9.
a=\frac{-\left(-12\right)±\sqrt{612}}{2\times 13}
Add 144 to 468.
a=\frac{-\left(-12\right)±6\sqrt{17}}{2\times 13}
Take the square root of 612.
a=\frac{12±6\sqrt{17}}{2\times 13}
The opposite of -12 is 12.
a=\frac{12±6\sqrt{17}}{26}
Multiply 2 times 13.
a=\frac{6\sqrt{17}+12}{26}
Now solve the equation a=\frac{12±6\sqrt{17}}{26} when ± is plus. Add 12 to 6\sqrt{17}.
a=\frac{3\sqrt{17}+6}{13}
Divide 12+6\sqrt{17} by 26.
a=\frac{12-6\sqrt{17}}{26}
Now solve the equation a=\frac{12±6\sqrt{17}}{26} when ± is minus. Subtract 6\sqrt{17} from 12.
a=\frac{6-3\sqrt{17}}{13}
Divide 12-6\sqrt{17} by 26.
a=\frac{3\sqrt{17}+6}{13} a=\frac{6-3\sqrt{17}}{13}
The equation is now solved.
13a^{2}-12a-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.
13a^{2}-12a-9-\left(-9\right)=-\left(-9\right)
Add 9 to both sides of the equation.
13a^{2}-12a=-\left(-9\right)
Subtracting -9 from itself leaves 0.
13a^{2}-12a=9
Subtract -9 from 0.
\frac{13a^{2}-12a}{13}=\frac{9}{13}
Divide both sides by 13.
a^{2}-\frac{12}{13}a=\frac{9}{13}
Dividing by 13 undoes the multiplication by 13.
a^{2}-\frac{12}{13}a+\left(-\frac{6}{13}\right)^{2}=\frac{9}{13}+\left(-\frac{6}{13}\right)^{2}
Divide -\frac{12}{13}, the coefficient of the x term, by 2 to get -\frac{6}{13}. Then add the square of -\frac{6}{13} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
a^{2}-\frac{12}{13}a+\frac{36}{169}=\frac{9}{13}+\frac{36}{169}
Square -\frac{6}{13} by squaring both the numerator and the denominator of the fraction.
a^{2}-\frac{12}{13}a+\frac{36}{169}=\frac{153}{169}
Add \frac{9}{13} to \frac{36}{169} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(a-\frac{6}{13}\right)^{2}=\frac{153}{169}
Factor a^{2}-\frac{12}{13}a+\frac{36}{169}. 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(a-\frac{6}{13}\right)^{2}}=\sqrt{\frac{153}{169}}
Take the square root of both sides of the equation.
a-\frac{6}{13}=\frac{3\sqrt{17}}{13} a-\frac{6}{13}=-\frac{3\sqrt{17}}{13}
Simplify.
a=\frac{3\sqrt{17}+6}{13} a=\frac{6-3\sqrt{17}}{13}
Add \frac{6}{13} to both sides of the equation.
x ^ 2 -\frac{12}{13}x -\frac{9}{13} = 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 13
r + s = \frac{12}{13} rs = -\frac{9}{13}
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{6}{13} - u s = \frac{6}{13} + u
Two numbers r and s sum up to \frac{12}{13} exactly when the average of the two numbers is \frac{1}{2}*\frac{12}{13} = \frac{6}{13}. 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{6}{13} - u) (\frac{6}{13} + u) = -\frac{9}{13}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{9}{13}
\frac{36}{169} - u^2 = -\frac{9}{13}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{9}{13}-\frac{36}{169} = -\frac{153}{169}
Simplify the expression by subtracting \frac{36}{169} on both sides
u^2 = \frac{153}{169} u = \pm\sqrt{\frac{153}{169}} = \pm \frac{\sqrt{153}}{13}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{6}{13} - \frac{\sqrt{153}}{13} = -0.490 s = \frac{6}{13} + \frac{\sqrt{153}}{13} = 1.413
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.