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a+b=-1 ab=-132
To solve the equation, factor n^{2}-n-132 using formula n^{2}+\left(a+b\right)n+ab=\left(n+a\right)\left(n+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 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 -132.
1-132=-131 2-66=-64 3-44=-41 4-33=-29 6-22=-16 11-12=-1
Calculate the sum for each pair.
a=-12 b=11
The solution is the pair that gives sum -1.
\left(n-12\right)\left(n+11\right)
Rewrite factored expression \left(n+a\right)\left(n+b\right) using the obtained values.
n=12 n=-11
To find equation solutions, solve n-12=0 and n+11=0.
a+b=-1 ab=1\left(-132\right)=-132
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as n^{2}+an+bn-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 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 -132.
1-132=-131 2-66=-64 3-44=-41 4-33=-29 6-22=-16 11-12=-1
Calculate the sum for each pair.
a=-12 b=11
The solution is the pair that gives sum -1.
\left(n^{2}-12n\right)+\left(11n-132\right)
Rewrite n^{2}-n-132 as \left(n^{2}-12n\right)+\left(11n-132\right).
n\left(n-12\right)+11\left(n-12\right)
Factor out n in the first and 11 in the second group.
\left(n-12\right)\left(n+11\right)
Factor out common term n-12 by using distributive property.
n=12 n=-11
To find equation solutions, solve n-12=0 and n+11=0.
n^{2}-n-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.
n=\frac{-\left(-1\right)±\sqrt{1-4\left(-132\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -1 for b, and -132 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
n=\frac{-\left(-1\right)±\sqrt{1+528}}{2}
Multiply -4 times -132.
n=\frac{-\left(-1\right)±\sqrt{529}}{2}
Add 1 to 528.
n=\frac{-\left(-1\right)±23}{2}
Take the square root of 529.
n=\frac{1±23}{2}
The opposite of -1 is 1.
n=\frac{24}{2}
Now solve the equation n=\frac{1±23}{2} when ± is plus. Add 1 to 23.
n=12
Divide 24 by 2.
n=-\frac{22}{2}
Now solve the equation n=\frac{1±23}{2} when ± is minus. Subtract 23 from 1.
n=-11
Divide -22 by 2.
n=12 n=-11
The equation is now solved.
n^{2}-n-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.
n^{2}-n-132-\left(-132\right)=-\left(-132\right)
Add 132 to both sides of the equation.
n^{2}-n=-\left(-132\right)
Subtracting -132 from itself leaves 0.
n^{2}-n=132
Subtract -132 from 0.
n^{2}-n+\left(-\frac{1}{2}\right)^{2}=132+\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.
n^{2}-n+\frac{1}{4}=132+\frac{1}{4}
Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.
n^{2}-n+\frac{1}{4}=\frac{529}{4}
Add 132 to \frac{1}{4}.
\left(n-\frac{1}{2}\right)^{2}=\frac{529}{4}
Factor n^{2}-n+\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(n-\frac{1}{2}\right)^{2}}=\sqrt{\frac{529}{4}}
Take the square root of both sides of the equation.
n-\frac{1}{2}=\frac{23}{2} n-\frac{1}{2}=-\frac{23}{2}
Simplify.
n=12 n=-11
Add \frac{1}{2} to both sides of the equation.
x ^ 2 -1x -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 = 1 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{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) = -132
To solve for unknown quantity u, substitute these in the product equation rs = -132
\frac{1}{4} - u^2 = -132
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -132-\frac{1}{4} = -\frac{529}{4}
Simplify the expression by subtracting \frac{1}{4} on both sides
u^2 = \frac{529}{4} u = \pm\sqrt{\frac{529}{4}} = \pm \frac{23}{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{1}{2} - \frac{23}{2} = -11 s = \frac{1}{2} + \frac{23}{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.