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k^{2}+9k+24-6=0
Subtract 6 from both sides.
k^{2}+9k+18=0
Subtract 6 from 24 to get 18.
a+b=9 ab=18
To solve the equation, factor k^{2}+9k+18 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.
1,18 2,9 3,6
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 18.
1+18=19 2+9=11 3+6=9
Calculate the sum for each pair.
a=3 b=6
The solution is the pair that gives sum 9.
\left(k+3\right)\left(k+6\right)
Rewrite factored expression \left(k+a\right)\left(k+b\right) using the obtained values.
k=-3 k=-6
To find equation solutions, solve k+3=0 and k+6=0.
k^{2}+9k+24-6=0
Subtract 6 from both sides.
k^{2}+9k+18=0
Subtract 6 from 24 to get 18.
a+b=9 ab=1\times 18=18
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as k^{2}+ak+bk+18. To find a and b, set up a system to be solved.
1,18 2,9 3,6
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 18.
1+18=19 2+9=11 3+6=9
Calculate the sum for each pair.
a=3 b=6
The solution is the pair that gives sum 9.
\left(k^{2}+3k\right)+\left(6k+18\right)
Rewrite k^{2}+9k+18 as \left(k^{2}+3k\right)+\left(6k+18\right).
k\left(k+3\right)+6\left(k+3\right)
Factor out k in the first and 6 in the second group.
\left(k+3\right)\left(k+6\right)
Factor out common term k+3 by using distributive property.
k=-3 k=-6
To find equation solutions, solve k+3=0 and k+6=0.
k^{2}+9k+24=6
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}+9k+24-6=6-6
Subtract 6 from both sides of the equation.
k^{2}+9k+24-6=0
Subtracting 6 from itself leaves 0.
k^{2}+9k+18=0
Subtract 6 from 24.
k=\frac{-9±\sqrt{9^{2}-4\times 18}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 9 for b, and 18 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
k=\frac{-9±\sqrt{81-4\times 18}}{2}
Square 9.
k=\frac{-9±\sqrt{81-72}}{2}
Multiply -4 times 18.
k=\frac{-9±\sqrt{9}}{2}
Add 81 to -72.
k=\frac{-9±3}{2}
Take the square root of 9.
k=-\frac{6}{2}
Now solve the equation k=\frac{-9±3}{2} when ± is plus. Add -9 to 3.
k=-3
Divide -6 by 2.
k=-\frac{12}{2}
Now solve the equation k=\frac{-9±3}{2} when ± is minus. Subtract 3 from -9.
k=-6
Divide -12 by 2.
k=-3 k=-6
The equation is now solved.
k^{2}+9k+24=6
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}+9k+24-24=6-24
Subtract 24 from both sides of the equation.
k^{2}+9k=6-24
Subtracting 24 from itself leaves 0.
k^{2}+9k=-18
Subtract 24 from 6.
k^{2}+9k+\left(\frac{9}{2}\right)^{2}=-18+\left(\frac{9}{2}\right)^{2}
Divide 9, the coefficient of the x term, by 2 to get \frac{9}{2}. Then add the square of \frac{9}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
k^{2}+9k+\frac{81}{4}=-18+\frac{81}{4}
Square \frac{9}{2} by squaring both the numerator and the denominator of the fraction.
k^{2}+9k+\frac{81}{4}=\frac{9}{4}
Add -18 to \frac{81}{4}.
\left(k+\frac{9}{2}\right)^{2}=\frac{9}{4}
Factor k^{2}+9k+\frac{81}{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{9}{2}\right)^{2}}=\sqrt{\frac{9}{4}}
Take the square root of both sides of the equation.
k+\frac{9}{2}=\frac{3}{2} k+\frac{9}{2}=-\frac{3}{2}
Simplify.
k=-3 k=-6
Subtract \frac{9}{2} from both sides of the equation.