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a+b=-9 ab=9\times 2=18
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 9c^{2}+ac+bc+2. To find a and b, set up a system to be solved.
-1,-18 -2,-9 -3,-6
Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 18.
-1-18=-19 -2-9=-11 -3-6=-9
Calculate the sum for each pair.
a=-6 b=-3
The solution is the pair that gives sum -9.
\left(9c^{2}-6c\right)+\left(-3c+2\right)
Rewrite 9c^{2}-9c+2 as \left(9c^{2}-6c\right)+\left(-3c+2\right).
3c\left(3c-2\right)-\left(3c-2\right)
Factor out 3c in the first and -1 in the second group.
\left(3c-2\right)\left(3c-1\right)
Factor out common term 3c-2 by using distributive property.
c=\frac{2}{3} c=\frac{1}{3}
To find equation solutions, solve 3c-2=0 and 3c-1=0.
9c^{2}-9c+2=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.
c=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 9\times 2}}{2\times 9}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, -9 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
c=\frac{-\left(-9\right)±\sqrt{81-4\times 9\times 2}}{2\times 9}
Square -9.
c=\frac{-\left(-9\right)±\sqrt{81-36\times 2}}{2\times 9}
Multiply -4 times 9.
c=\frac{-\left(-9\right)±\sqrt{81-72}}{2\times 9}
Multiply -36 times 2.
c=\frac{-\left(-9\right)±\sqrt{9}}{2\times 9}
Add 81 to -72.
c=\frac{-\left(-9\right)±3}{2\times 9}
Take the square root of 9.
c=\frac{9±3}{2\times 9}
The opposite of -9 is 9.
c=\frac{9±3}{18}
Multiply 2 times 9.
c=\frac{12}{18}
Now solve the equation c=\frac{9±3}{18} when ± is plus. Add 9 to 3.
c=\frac{2}{3}
Reduce the fraction \frac{12}{18} to lowest terms by extracting and canceling out 6.
c=\frac{6}{18}
Now solve the equation c=\frac{9±3}{18} when ± is minus. Subtract 3 from 9.
c=\frac{1}{3}
Reduce the fraction \frac{6}{18} to lowest terms by extracting and canceling out 6.
c=\frac{2}{3} c=\frac{1}{3}
The equation is now solved.
9c^{2}-9c+2=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.
9c^{2}-9c+2-2=-2
Subtract 2 from both sides of the equation.
9c^{2}-9c=-2
Subtracting 2 from itself leaves 0.
\frac{9c^{2}-9c}{9}=-\frac{2}{9}
Divide both sides by 9.
c^{2}+\left(-\frac{9}{9}\right)c=-\frac{2}{9}
Dividing by 9 undoes the multiplication by 9.
c^{2}-c=-\frac{2}{9}
Divide -9 by 9.
c^{2}-c+\left(-\frac{1}{2}\right)^{2}=-\frac{2}{9}+\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.
c^{2}-c+\frac{1}{4}=-\frac{2}{9}+\frac{1}{4}
Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.
c^{2}-c+\frac{1}{4}=\frac{1}{36}
Add -\frac{2}{9} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(c-\frac{1}{2}\right)^{2}=\frac{1}{36}
Factor c^{2}-c+\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(c-\frac{1}{2}\right)^{2}}=\sqrt{\frac{1}{36}}
Take the square root of both sides of the equation.
c-\frac{1}{2}=\frac{1}{6} c-\frac{1}{2}=-\frac{1}{6}
Simplify.
c=\frac{2}{3} c=\frac{1}{3}
Add \frac{1}{2} to both sides of the equation.
x ^ 2 -1x +\frac{2}{9} = 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 = 1 rs = \frac{2}{9}
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{2}{9}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{2}{9}
\frac{1}{4} - u^2 = \frac{2}{9}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{2}{9}-\frac{1}{4} = -\frac{1}{36}
Simplify the expression by subtracting \frac{1}{4} on both sides
u^2 = \frac{1}{36} u = \pm\sqrt{\frac{1}{36}} = \pm \frac{1}{6}
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}{6} = 0.333 s = \frac{1}{2} + \frac{1}{6} = 0.667
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.