Solve for x
x = -\frac{8}{3} = -2\frac{2}{3} \approx -2.666666667
x = \frac{4}{3} = 1\frac{1}{3} \approx 1.333333333
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a+b=12 ab=9\left(-32\right)=-288
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 9x^{2}+ax+bx-32. To find a and b, set up a system to be solved.
-1,288 -2,144 -3,96 -4,72 -6,48 -8,36 -9,32 -12,24 -16,18
Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -288.
-1+288=287 -2+144=142 -3+96=93 -4+72=68 -6+48=42 -8+36=28 -9+32=23 -12+24=12 -16+18=2
Calculate the sum for each pair.
a=-12 b=24
The solution is the pair that gives sum 12.
\left(9x^{2}-12x\right)+\left(24x-32\right)
Rewrite 9x^{2}+12x-32 as \left(9x^{2}-12x\right)+\left(24x-32\right).
3x\left(3x-4\right)+8\left(3x-4\right)
Factor out 3x in the first and 8 in the second group.
\left(3x-4\right)\left(3x+8\right)
Factor out common term 3x-4 by using distributive property.
x=\frac{4}{3} x=-\frac{8}{3}
To find equation solutions, solve 3x-4=0 and 3x+8=0.
9x^{2}+12x-32=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{-12±\sqrt{12^{2}-4\times 9\left(-32\right)}}{2\times 9}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, 12 for b, and -32 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-12±\sqrt{144-4\times 9\left(-32\right)}}{2\times 9}
Square 12.
x=\frac{-12±\sqrt{144-36\left(-32\right)}}{2\times 9}
Multiply -4 times 9.
x=\frac{-12±\sqrt{144+1152}}{2\times 9}
Multiply -36 times -32.
x=\frac{-12±\sqrt{1296}}{2\times 9}
Add 144 to 1152.
x=\frac{-12±36}{2\times 9}
Take the square root of 1296.
x=\frac{-12±36}{18}
Multiply 2 times 9.
x=\frac{24}{18}
Now solve the equation x=\frac{-12±36}{18} when ± is plus. Add -12 to 36.
x=\frac{4}{3}
Reduce the fraction \frac{24}{18} to lowest terms by extracting and canceling out 6.
x=-\frac{48}{18}
Now solve the equation x=\frac{-12±36}{18} when ± is minus. Subtract 36 from -12.
x=-\frac{8}{3}
Reduce the fraction \frac{-48}{18} to lowest terms by extracting and canceling out 6.
x=\frac{4}{3} x=-\frac{8}{3}
The equation is now solved.
9x^{2}+12x-32=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}+12x-32-\left(-32\right)=-\left(-32\right)
Add 32 to both sides of the equation.
9x^{2}+12x=-\left(-32\right)
Subtracting -32 from itself leaves 0.
9x^{2}+12x=32
Subtract -32 from 0.
\frac{9x^{2}+12x}{9}=\frac{32}{9}
Divide both sides by 9.
x^{2}+\frac{12}{9}x=\frac{32}{9}
Dividing by 9 undoes the multiplication by 9.
x^{2}+\frac{4}{3}x=\frac{32}{9}
Reduce the fraction \frac{12}{9} to lowest terms by extracting and canceling out 3.
x^{2}+\frac{4}{3}x+\left(\frac{2}{3}\right)^{2}=\frac{32}{9}+\left(\frac{2}{3}\right)^{2}
Divide \frac{4}{3}, the coefficient of the x term, by 2 to get \frac{2}{3}. Then add the square of \frac{2}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+\frac{4}{3}x+\frac{4}{9}=\frac{32+4}{9}
Square \frac{2}{3} by squaring both the numerator and the denominator of the fraction.
x^{2}+\frac{4}{3}x+\frac{4}{9}=4
Add \frac{32}{9} to \frac{4}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x+\frac{2}{3}\right)^{2}=4
Factor x^{2}+\frac{4}{3}x+\frac{4}{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(x+\frac{2}{3}\right)^{2}}=\sqrt{4}
Take the square root of both sides of the equation.
x+\frac{2}{3}=2 x+\frac{2}{3}=-2
Simplify.
x=\frac{4}{3} x=-\frac{8}{3}
Subtract \frac{2}{3} from both sides of the equation.
x ^ 2 +\frac{4}{3}x -\frac{32}{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{4}{3} rs = -\frac{32}{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{2}{3} - u s = -\frac{2}{3} + u
Two numbers r and s sum up to -\frac{4}{3} exactly when the average of the two numbers is \frac{1}{2}*-\frac{4}{3} = -\frac{2}{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{2}{3} - u) (-\frac{2}{3} + u) = -\frac{32}{9}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{32}{9}
\frac{4}{9} - u^2 = -\frac{32}{9}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{32}{9}-\frac{4}{9} = -4
Simplify the expression by subtracting \frac{4}{9} on both sides
u^2 = 4 u = \pm\sqrt{4} = \pm 2
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-\frac{2}{3} - 2 = -2.667 s = -\frac{2}{3} + 2 = 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.
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