Solve for x
x=11
x=12
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a+b=-23 ab=132
To solve the equation, factor x^{2}-23x+132 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.
-1,-132 -2,-66 -3,-44 -4,-33 -6,-22 -11,-12
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 132.
-1-132=-133 -2-66=-68 -3-44=-47 -4-33=-37 -6-22=-28 -11-12=-23
Calculate the sum for each pair.
a=-12 b=-11
The solution is the pair that gives sum -23.
\left(x-12\right)\left(x-11\right)
Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.
x=12 x=11
To find equation solutions, solve x-12=0 and x-11=0.
a+b=-23 ab=1\times 132=132
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+132. To find a and b, set up a system to be solved.
-1,-132 -2,-66 -3,-44 -4,-33 -6,-22 -11,-12
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 132.
-1-132=-133 -2-66=-68 -3-44=-47 -4-33=-37 -6-22=-28 -11-12=-23
Calculate the sum for each pair.
a=-12 b=-11
The solution is the pair that gives sum -23.
\left(x^{2}-12x\right)+\left(-11x+132\right)
Rewrite x^{2}-23x+132 as \left(x^{2}-12x\right)+\left(-11x+132\right).
x\left(x-12\right)-11\left(x-12\right)
Factor out x in the first and -11 in the second group.
\left(x-12\right)\left(x-11\right)
Factor out common term x-12 by using distributive property.
x=12 x=11
To find equation solutions, solve x-12=0 and x-11=0.
x^{2}-23x+132=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(-23\right)±\sqrt{\left(-23\right)^{2}-4\times 132}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -23 for b, and 132 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-23\right)±\sqrt{529-4\times 132}}{2}
Square -23.
x=\frac{-\left(-23\right)±\sqrt{529-528}}{2}
Multiply -4 times 132.
x=\frac{-\left(-23\right)±\sqrt{1}}{2}
Add 529 to -528.
x=\frac{-\left(-23\right)±1}{2}
Take the square root of 1.
x=\frac{23±1}{2}
The opposite of -23 is 23.
x=\frac{24}{2}
Now solve the equation x=\frac{23±1}{2} when ± is plus. Add 23 to 1.
x=12
Divide 24 by 2.
x=\frac{22}{2}
Now solve the equation x=\frac{23±1}{2} when ± is minus. Subtract 1 from 23.
x=11
Divide 22 by 2.
x=12 x=11
The equation is now solved.
x^{2}-23x+132=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.
x^{2}-23x+132-132=-132
Subtract 132 from both sides of the equation.
x^{2}-23x=-132
Subtracting 132 from itself leaves 0.
x^{2}-23x+\left(-\frac{23}{2}\right)^{2}=-132+\left(-\frac{23}{2}\right)^{2}
Divide -23, the coefficient of the x term, by 2 to get -\frac{23}{2}. Then add the square of -\frac{23}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-23x+\frac{529}{4}=-132+\frac{529}{4}
Square -\frac{23}{2} by squaring both the numerator and the denominator of the fraction.
x^{2}-23x+\frac{529}{4}=\frac{1}{4}
Add -132 to \frac{529}{4}.
\left(x-\frac{23}{2}\right)^{2}=\frac{1}{4}
Factor x^{2}-23x+\frac{529}{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(x-\frac{23}{2}\right)^{2}}=\sqrt{\frac{1}{4}}
Take the square root of both sides of the equation.
x-\frac{23}{2}=\frac{1}{2} x-\frac{23}{2}=-\frac{1}{2}
Simplify.
x=12 x=11
Add \frac{23}{2} to both sides of the equation.
x ^ 2 -23x +132 = 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.
r + s = 23 rs = 132
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{23}{2} - u s = \frac{23}{2} + u
Two numbers r and s sum up to 23 exactly when the average of the two numbers is \frac{1}{2}*23 = \frac{23}{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{23}{2} - u) (\frac{23}{2} + u) = 132
To solve for unknown quantity u, substitute these in the product equation rs = 132
\frac{529}{4} - u^2 = 132
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 132-\frac{529}{4} = -\frac{1}{4}
Simplify the expression by subtracting \frac{529}{4} on both sides
u^2 = \frac{1}{4} u = \pm\sqrt{\frac{1}{4}} = \pm \frac{1}{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{23}{2} - \frac{1}{2} = 11 s = \frac{23}{2} + \frac{1}{2} = 12
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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