Solve for n
n=-22
n=11
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n^{2}+11n-242=0
Subtract 242 from both sides.
a+b=11 ab=-242
To solve the equation, factor n^{2}+11n-242 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.
-1,242 -2,121 -11,22
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 -242.
-1+242=241 -2+121=119 -11+22=11
Calculate the sum for each pair.
a=-11 b=22
The solution is the pair that gives sum 11.
\left(n-11\right)\left(n+22\right)
Rewrite factored expression \left(n+a\right)\left(n+b\right) using the obtained values.
n=11 n=-22
To find equation solutions, solve n-11=0 and n+22=0.
n^{2}+11n-242=0
Subtract 242 from both sides.
a+b=11 ab=1\left(-242\right)=-242
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as n^{2}+an+bn-242. To find a and b, set up a system to be solved.
-1,242 -2,121 -11,22
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 -242.
-1+242=241 -2+121=119 -11+22=11
Calculate the sum for each pair.
a=-11 b=22
The solution is the pair that gives sum 11.
\left(n^{2}-11n\right)+\left(22n-242\right)
Rewrite n^{2}+11n-242 as \left(n^{2}-11n\right)+\left(22n-242\right).
n\left(n-11\right)+22\left(n-11\right)
Factor out n in the first and 22 in the second group.
\left(n-11\right)\left(n+22\right)
Factor out common term n-11 by using distributive property.
n=11 n=-22
To find equation solutions, solve n-11=0 and n+22=0.
n^{2}+11n=242
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^{2}+11n-242=242-242
Subtract 242 from both sides of the equation.
n^{2}+11n-242=0
Subtracting 242 from itself leaves 0.
n=\frac{-11±\sqrt{11^{2}-4\left(-242\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 11 for b, and -242 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
n=\frac{-11±\sqrt{121-4\left(-242\right)}}{2}
Square 11.
n=\frac{-11±\sqrt{121+968}}{2}
Multiply -4 times -242.
n=\frac{-11±\sqrt{1089}}{2}
Add 121 to 968.
n=\frac{-11±33}{2}
Take the square root of 1089.
n=\frac{22}{2}
Now solve the equation n=\frac{-11±33}{2} when ± is plus. Add -11 to 33.
n=11
Divide 22 by 2.
n=-\frac{44}{2}
Now solve the equation n=\frac{-11±33}{2} when ± is minus. Subtract 33 from -11.
n=-22
Divide -44 by 2.
n=11 n=-22
The equation is now solved.
n^{2}+11n=242
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}+11n+\left(\frac{11}{2}\right)^{2}=242+\left(\frac{11}{2}\right)^{2}
Divide 11, the coefficient of the x term, by 2 to get \frac{11}{2}. Then add the square of \frac{11}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
n^{2}+11n+\frac{121}{4}=242+\frac{121}{4}
Square \frac{11}{2} by squaring both the numerator and the denominator of the fraction.
n^{2}+11n+\frac{121}{4}=\frac{1089}{4}
Add 242 to \frac{121}{4}.
\left(n+\frac{11}{2}\right)^{2}=\frac{1089}{4}
Factor n^{2}+11n+\frac{121}{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{11}{2}\right)^{2}}=\sqrt{\frac{1089}{4}}
Take the square root of both sides of the equation.
n+\frac{11}{2}=\frac{33}{2} n+\frac{11}{2}=-\frac{33}{2}
Simplify.
n=11 n=-22
Subtract \frac{11}{2} from both sides of the equation.
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