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k^{2}+15k+56=0
Add 56 to both sides.
a+b=15 ab=56
To solve the equation, factor k^{2}+15k+56 using formula k^{2}+\left(a+b\right)k+ab=\left(k+a\right)\left(k+b\right). To find a and b, set up a system to be solved.
1,56 2,28 4,14 7,8
Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 56.
1+56=57 2+28=30 4+14=18 7+8=15
Calculate the sum for each pair.
a=7 b=8
The solution is the pair that gives sum 15.
\left(k+7\right)\left(k+8\right)
Rewrite factored expression \left(k+a\right)\left(k+b\right) using the obtained values.
k=-7 k=-8
To find equation solutions, solve k+7=0 and k+8=0.
k^{2}+15k+56=0
Add 56 to both sides.
a+b=15 ab=1\times 56=56
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as k^{2}+ak+bk+56. To find a and b, set up a system to be solved.
1,56 2,28 4,14 7,8
Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 56.
1+56=57 2+28=30 4+14=18 7+8=15
Calculate the sum for each pair.
a=7 b=8
The solution is the pair that gives sum 15.
\left(k^{2}+7k\right)+\left(8k+56\right)
Rewrite k^{2}+15k+56 as \left(k^{2}+7k\right)+\left(8k+56\right).
k\left(k+7\right)+8\left(k+7\right)
Factor out k in the first and 8 in the second group.
\left(k+7\right)\left(k+8\right)
Factor out common term k+7 by using distributive property.
k=-7 k=-8
To find equation solutions, solve k+7=0 and k+8=0.
k^{2}+15k=-56
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.
k^{2}+15k-\left(-56\right)=-56-\left(-56\right)
Add 56 to both sides of the equation.
k^{2}+15k-\left(-56\right)=0
Subtracting -56 from itself leaves 0.
k^{2}+15k+56=0
Subtract -56 from 0.
k=\frac{-15±\sqrt{15^{2}-4\times 56}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 15 for b, and 56 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
k=\frac{-15±\sqrt{225-4\times 56}}{2}
Square 15.
k=\frac{-15±\sqrt{225-224}}{2}
Multiply -4 times 56.
k=\frac{-15±\sqrt{1}}{2}
Add 225 to -224.
k=\frac{-15±1}{2}
Take the square root of 1.
k=-\frac{14}{2}
Now solve the equation k=\frac{-15±1}{2} when ± is plus. Add -15 to 1.
k=-7
Divide -14 by 2.
k=-\frac{16}{2}
Now solve the equation k=\frac{-15±1}{2} when ± is minus. Subtract 1 from -15.
k=-8
Divide -16 by 2.
k=-7 k=-8
The equation is now solved.
k^{2}+15k=-56
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.
k^{2}+15k+\left(\frac{15}{2}\right)^{2}=-56+\left(\frac{15}{2}\right)^{2}
Divide 15, the coefficient of the x term, by 2 to get \frac{15}{2}. Then add the square of \frac{15}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
k^{2}+15k+\frac{225}{4}=-56+\frac{225}{4}
Square \frac{15}{2} by squaring both the numerator and the denominator of the fraction.
k^{2}+15k+\frac{225}{4}=\frac{1}{4}
Add -56 to \frac{225}{4}.
\left(k+\frac{15}{2}\right)^{2}=\frac{1}{4}
Factor k^{2}+15k+\frac{225}{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(k+\frac{15}{2}\right)^{2}}=\sqrt{\frac{1}{4}}
Take the square root of both sides of the equation.
k+\frac{15}{2}=\frac{1}{2} k+\frac{15}{2}=-\frac{1}{2}
Simplify.
k=-7 k=-8
Subtract \frac{15}{2} from both sides of the equation.