Solve for k

Steps Using Factoring By Grouping
Steps Using the Quadratic Formula
Steps for Completing the Square
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2k^{2}+9k+7=0
Add 7 to both sides.
a+b=9 ab=2\times 7=14
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2k^{2}+ak+bk+7. To find a and b, set up a system to be solved.
1,14 2,7
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 14.
1+14=15 2+7=9
Calculate the sum for each pair.
a=2 b=7
The solution is the pair that gives sum 9.
\left(2k^{2}+2k\right)+\left(7k+7\right)
Rewrite 2k^{2}+9k+7 as \left(2k^{2}+2k\right)+\left(7k+7\right).
2k\left(k+1\right)+7\left(k+1\right)
Factor out 2k in the first and 7 in the second group.
\left(k+1\right)\left(2k+7\right)
Factor out common term k+1 by using distributive property.
k=-1 k=-\frac{7}{2}
To find equation solutions, solve k+1=0 and 2k+7=0.
2k^{2}+9k=-7
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.
2k^{2}+9k-\left(-7\right)=-7-\left(-7\right)
Add 7 to both sides of the equation.
2k^{2}+9k-\left(-7\right)=0
Subtracting -7 from itself leaves 0.
2k^{2}+9k+7=0
Subtract -7 from 0.
k=\frac{-9±\sqrt{9^{2}-4\times 2\times 7}}{2\times 2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 9 for b, and 7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
k=\frac{-9±\sqrt{81-4\times 2\times 7}}{2\times 2}
Square 9.
k=\frac{-9±\sqrt{81-8\times 7}}{2\times 2}
Multiply -4 times 2.
k=\frac{-9±\sqrt{81-56}}{2\times 2}
Multiply -8 times 7.
k=\frac{-9±\sqrt{25}}{2\times 2}
Add 81 to -56.
k=\frac{-9±5}{2\times 2}
Take the square root of 25.
k=\frac{-9±5}{4}
Multiply 2 times 2.
k=\frac{-4}{4}
Now solve the equation k=\frac{-9±5}{4} when ± is plus. Add -9 to 5.
k=-1
Divide -4 by 4.
k=\frac{-14}{4}
Now solve the equation k=\frac{-9±5}{4} when ± is minus. Subtract 5 from -9.
k=-\frac{7}{2}
Reduce the fraction \frac{-14}{4}=-3.5 to lowest terms by extracting and canceling out 2.
k=-1 k=-\frac{7}{2}
The equation is now solved.
2k^{2}+9k=-7
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.
\frac{2k^{2}+9k}{2}=\frac{-7}{2}
Divide both sides by 2.
k^{2}+\frac{9}{2}k=\frac{-7}{2}
Dividing by 2 undoes the multiplication by 2.
k^{2}+\frac{9}{2}k=-\frac{7}{2}
Divide -7 by 2.
k^{2}+\frac{9}{2}k+\left(\frac{9}{4}\right)^{2}=-\frac{7}{2}+\left(\frac{9}{4}\right)^{2}
Divide \frac{9}{2}=4.5, the coefficient of the x term, by 2 to get \frac{9}{4}=2.25. Then add the square of \frac{9}{4}=2.25 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
k^{2}+\frac{9}{2}k+\frac{81}{16}=-\frac{7}{2}+\frac{81}{16}
Square \frac{9}{4}=2.25 by squaring both the numerator and the denominator of the fraction.
k^{2}+\frac{9}{2}k+\frac{81}{16}=\frac{25}{16}
Add -\frac{7}{2}=-3.5 to \frac{81}{16}=5.0625 by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(k+\frac{9}{4}\right)^{2}=\frac{25}{16}
Factor k^{2}+\frac{9}{2}k+\frac{81}{16}. 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}{4}\right)^{2}}=\sqrt{\frac{25}{16}}
Take the square root of both sides of the equation.
k+\frac{9}{4}=\frac{5}{4} k+\frac{9}{4}=-\frac{5}{4}
Simplify.
k=-1 k=-\frac{7}{2}
Subtract \frac{9}{4}=2.25 from both sides of the equation.