Solve for x
x=-\frac{2}{9}\approx -0.222222222
x=18
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a+b=-160 ab=9\left(-36\right)=-324
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 9x^{2}+ax+bx-36. To find a and b, set up a system to be solved.
1,-324 2,-162 3,-108 4,-81 6,-54 9,-36 12,-27 18,-18
Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -324.
1-324=-323 2-162=-160 3-108=-105 4-81=-77 6-54=-48 9-36=-27 12-27=-15 18-18=0
Calculate the sum for each pair.
a=-162 b=2
The solution is the pair that gives sum -160.
\left(9x^{2}-162x\right)+\left(2x-36\right)
Rewrite 9x^{2}-160x-36 as \left(9x^{2}-162x\right)+\left(2x-36\right).
9x\left(x-18\right)+2\left(x-18\right)
Factor out 9x in the first and 2 in the second group.
\left(x-18\right)\left(9x+2\right)
Factor out common term x-18 by using distributive property.
x=18 x=-\frac{2}{9}
To find equation solutions, solve x-18=0 and 9x+2=0.
9x^{2}-160x-36=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{-\left(-160\right)±\sqrt{\left(-160\right)^{2}-4\times 9\left(-36\right)}}{2\times 9}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, -160 for b, and -36 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-160\right)±\sqrt{25600-4\times 9\left(-36\right)}}{2\times 9}
Square -160.
x=\frac{-\left(-160\right)±\sqrt{25600-36\left(-36\right)}}{2\times 9}
Multiply -4 times 9.
x=\frac{-\left(-160\right)±\sqrt{25600+1296}}{2\times 9}
Multiply -36 times -36.
x=\frac{-\left(-160\right)±\sqrt{26896}}{2\times 9}
Add 25600 to 1296.
x=\frac{-\left(-160\right)±164}{2\times 9}
Take the square root of 26896.
x=\frac{160±164}{2\times 9}
The opposite of -160 is 160.
x=\frac{160±164}{18}
Multiply 2 times 9.
x=\frac{324}{18}
Now solve the equation x=\frac{160±164}{18} when ± is plus. Add 160 to 164.
x=18
Divide 324 by 18.
x=-\frac{4}{18}
Now solve the equation x=\frac{160±164}{18} when ± is minus. Subtract 164 from 160.
x=-\frac{2}{9}
Reduce the fraction \frac{-4}{18} to lowest terms by extracting and canceling out 2.
x=18 x=-\frac{2}{9}
The equation is now solved.
9x^{2}-160x-36=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}-160x-36-\left(-36\right)=-\left(-36\right)
Add 36 to both sides of the equation.
9x^{2}-160x=-\left(-36\right)
Subtracting -36 from itself leaves 0.
9x^{2}-160x=36
Subtract -36 from 0.
\frac{9x^{2}-160x}{9}=\frac{36}{9}
Divide both sides by 9.
x^{2}-\frac{160}{9}x=\frac{36}{9}
Dividing by 9 undoes the multiplication by 9.
x^{2}-\frac{160}{9}x=4
Divide 36 by 9.
x^{2}-\frac{160}{9}x+\left(-\frac{80}{9}\right)^{2}=4+\left(-\frac{80}{9}\right)^{2}
Divide -\frac{160}{9}, the coefficient of the x term, by 2 to get -\frac{80}{9}. Then add the square of -\frac{80}{9} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{160}{9}x+\frac{6400}{81}=4+\frac{6400}{81}
Square -\frac{80}{9} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{160}{9}x+\frac{6400}{81}=\frac{6724}{81}
Add 4 to \frac{6400}{81}.
\left(x-\frac{80}{9}\right)^{2}=\frac{6724}{81}
Factor x^{2}-\frac{160}{9}x+\frac{6400}{81}. 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{80}{9}\right)^{2}}=\sqrt{\frac{6724}{81}}
Take the square root of both sides of the equation.
x-\frac{80}{9}=\frac{82}{9} x-\frac{80}{9}=-\frac{82}{9}
Simplify.
x=18 x=-\frac{2}{9}
Add \frac{80}{9} to both sides of the equation.
x ^ 2 -\frac{160}{9}x -4 = 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 9
r + s = \frac{160}{9} rs = -4
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{80}{9} - u s = \frac{80}{9} + u
Two numbers r and s sum up to \frac{160}{9} exactly when the average of the two numbers is \frac{1}{2}*\frac{160}{9} = \frac{80}{9}. 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{80}{9} - u) (\frac{80}{9} + u) = -4
To solve for unknown quantity u, substitute these in the product equation rs = -4
\frac{6400}{81} - u^2 = -4
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -4-\frac{6400}{81} = -\frac{6724}{81}
Simplify the expression by subtracting \frac{6400}{81} on both sides
u^2 = \frac{6724}{81} u = \pm\sqrt{\frac{6724}{81}} = \pm \frac{82}{9}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{80}{9} - \frac{82}{9} = -0.222 s = \frac{80}{9} + \frac{82}{9} = 18
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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