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a+b=-6 ab=9\left(-8\right)=-72
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 9s^{2}+as+bs-8. To find a and b, set up a system to be solved.
1,-72 2,-36 3,-24 4,-18 6,-12 8,-9
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 -72.
1-72=-71 2-36=-34 3-24=-21 4-18=-14 6-12=-6 8-9=-1
Calculate the sum for each pair.
a=-12 b=6
The solution is the pair that gives sum -6.
\left(9s^{2}-12s\right)+\left(6s-8\right)
Rewrite 9s^{2}-6s-8 as \left(9s^{2}-12s\right)+\left(6s-8\right).
3s\left(3s-4\right)+2\left(3s-4\right)
Factor out 3s in the first and 2 in the second group.
\left(3s-4\right)\left(3s+2\right)
Factor out common term 3s-4 by using distributive property.
s=\frac{4}{3} s=-\frac{2}{3}
To find equation solutions, solve 3s-4=0 and 3s+2=0.
9s^{2}-6s-8=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.
s=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 9\left(-8\right)}}{2\times 9}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, -6 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
s=\frac{-\left(-6\right)±\sqrt{36-4\times 9\left(-8\right)}}{2\times 9}
Square -6.
s=\frac{-\left(-6\right)±\sqrt{36-36\left(-8\right)}}{2\times 9}
Multiply -4 times 9.
s=\frac{-\left(-6\right)±\sqrt{36+288}}{2\times 9}
Multiply -36 times -8.
s=\frac{-\left(-6\right)±\sqrt{324}}{2\times 9}
Add 36 to 288.
s=\frac{-\left(-6\right)±18}{2\times 9}
Take the square root of 324.
s=\frac{6±18}{2\times 9}
The opposite of -6 is 6.
s=\frac{6±18}{18}
Multiply 2 times 9.
s=\frac{24}{18}
Now solve the equation s=\frac{6±18}{18} when ± is plus. Add 6 to 18.
s=\frac{4}{3}
Reduce the fraction \frac{24}{18} to lowest terms by extracting and canceling out 6.
s=-\frac{12}{18}
Now solve the equation s=\frac{6±18}{18} when ± is minus. Subtract 18 from 6.
s=-\frac{2}{3}
Reduce the fraction \frac{-12}{18} to lowest terms by extracting and canceling out 6.
s=\frac{4}{3} s=-\frac{2}{3}
The equation is now solved.
9s^{2}-6s-8=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.
9s^{2}-6s-8-\left(-8\right)=-\left(-8\right)
Add 8 to both sides of the equation.
9s^{2}-6s=-\left(-8\right)
Subtracting -8 from itself leaves 0.
9s^{2}-6s=8
Subtract -8 from 0.
\frac{9s^{2}-6s}{9}=\frac{8}{9}
Divide both sides by 9.
s^{2}+\left(-\frac{6}{9}\right)s=\frac{8}{9}
Dividing by 9 undoes the multiplication by 9.
s^{2}-\frac{2}{3}s=\frac{8}{9}
Reduce the fraction \frac{-6}{9} to lowest terms by extracting and canceling out 3.
s^{2}-\frac{2}{3}s+\left(-\frac{1}{3}\right)^{2}=\frac{8}{9}+\left(-\frac{1}{3}\right)^{2}
Divide -\frac{2}{3}, the coefficient of the x term, by 2 to get -\frac{1}{3}. Then add the square of -\frac{1}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
s^{2}-\frac{2}{3}s+\frac{1}{9}=\frac{8+1}{9}
Square -\frac{1}{3} by squaring both the numerator and the denominator of the fraction.
s^{2}-\frac{2}{3}s+\frac{1}{9}=1
Add \frac{8}{9} to \frac{1}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(s-\frac{1}{3}\right)^{2}=1
Factor s^{2}-\frac{2}{3}s+\frac{1}{9}. 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(s-\frac{1}{3}\right)^{2}}=\sqrt{1}
Take the square root of both sides of the equation.
s-\frac{1}{3}=1 s-\frac{1}{3}=-1
Simplify.
s=\frac{4}{3} s=-\frac{2}{3}
Add \frac{1}{3} to both sides of the equation.
x ^ 2 -\frac{2}{3}x -\frac{8}{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 = \frac{2}{3} rs = -\frac{8}{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}{3} - u s = \frac{1}{3} + u
Two numbers r and s sum up to \frac{2}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{2}{3} = \frac{1}{3}. 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}{3} - u) (\frac{1}{3} + u) = -\frac{8}{9}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{8}{9}
\frac{1}{9} - u^2 = -\frac{8}{9}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{8}{9}-\frac{1}{9} = -1
Simplify the expression by subtracting \frac{1}{9} on both sides
u^2 = 1 u = \pm\sqrt{1} = \pm 1
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}{3} - 1 = -0.667 s = \frac{1}{3} + 1 = 1.333
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.