Solve 9+k=18-k^2/k

$$9 + k = \frac { 18 - k ^ { 2 } } { k }$$

Solve for k

$$k=-6$$

$$k = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

Steps Using Factoring By Grouping

Variable k cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by k.

$$k\times 9+kk=18-k^{2}$$

Multiply k and k to get k^{2}.

$$k\times 9+k^{2}=18-k^{2}$$

Subtract 18 from both sides.

$$k\times 9+k^{2}-18=-k^{2}$$

Add k^{2} to both sides.

$$k\times 9+k^{2}-18+k^{2}=0$$

Combine k^{2} and k^{2} to get 2k^{2}.

$$k\times 9+2k^{2}-18=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$2k^{2}+9k-18=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2k^{2}+ak+bk-18. To find a and b, set up a system to be solved.

$$a+b=9 ab=2\left(-18\right)=-36$$

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 -36.

$$-1,36 -2,18 -3,12 -4,9 -6,6$$

Calculate the sum for each pair.

$$-1+36=35 -2+18=16 -3+12=9 -4+9=5 -6+6=0$$

The solution is the pair that gives sum 9.

$$a=-3 b=12$$

Rewrite 2k^{2}+9k-18 as \left(2k^{2}-3k\right)+\left(12k-18\right).

$$\left(2k^{2}-3k\right)+\left(12k-18\right)$$

Factor out k in the first and 6 in the second group.

$$k\left(2k-3\right)+6\left(2k-3\right)$$

Factor out common term 2k-3 by using distributive property.

$$\left(2k-3\right)\left(k+6\right)$$

To find equation solutions, solve 2k-3=0 and k+6=0.

$$k=\frac{3}{2} k=-6$$

Steps Using the Quadratic Formula

Variable k cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by k.

$$k\times 9+kk=18-k^{2}$$

Multiply k and k to get k^{2}.

$$k\times 9+k^{2}=18-k^{2}$$

Subtract 18 from both sides.

$$k\times 9+k^{2}-18=-k^{2}$$

Add k^{2} to both sides.

$$k\times 9+k^{2}-18+k^{2}=0$$

Combine k^{2} and k^{2} to get 2k^{2}.

$$k\times 9+2k^{2}-18=0$$

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.

$$2k^{2}+9k-18=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 9 for b, and -18 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$k=\frac{-9±\sqrt{9^{2}-4\times 2\left(-18\right)}}{2\times 2}$$

Square 9.

$$k=\frac{-9±\sqrt{81-4\times 2\left(-18\right)}}{2\times 2}$$

Multiply -4 times 2.

$$k=\frac{-9±\sqrt{81-8\left(-18\right)}}{2\times 2}$$

Multiply -8 times -18.

$$k=\frac{-9±\sqrt{81+144}}{2\times 2}$$

Add 81 to 144.

$$k=\frac{-9±\sqrt{225}}{2\times 2}$$

Take the square root of 225.

$$k=\frac{-9±15}{2\times 2}$$

Multiply 2 times 2.

$$k=\frac{-9±15}{4}$$

Now solve the equation k=\frac{-9±15}{4} when ± is plus. Add -9 to 15.

$$k=\frac{6}{4}$$

Reduce the fraction \frac{6}{4} to lowest terms by extracting and canceling out 2.

$$k=\frac{3}{2}$$

Now solve the equation k=\frac{-9±15}{4} when ± is minus. Subtract 15 from -9.

$$k=-\frac{24}{4}$$

Divide -24 by 4.

$$k=-6$$

The equation is now solved.

$$k=\frac{3}{2} k=-6$$

Steps for Completing the Square

Variable k cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by k.

$$k\times 9+kk=18-k^{2}$$

Multiply k and k to get k^{2}.

$$k\times 9+k^{2}=18-k^{2}$$

Add k^{2} to both sides.

$$k\times 9+k^{2}+k^{2}=18$$

Combine k^{2} and k^{2} to get 2k^{2}.

$$k\times 9+2k^{2}=18$$

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.

$$2k^{2}+9k=18$$

Divide both sides by 2.

$$\frac{2k^{2}+9k}{2}=\frac{18}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$k^{2}+\frac{9}{2}k=\frac{18}{2}$$

Divide 18 by 2.

$$k^{2}+\frac{9}{2}k=9$$

Divide \frac{9}{2}, the coefficient of the x term, by 2 to get \frac{9}{4}. Then add the square of \frac{9}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$k^{2}+\frac{9}{2}k+\left(\frac{9}{4}\right)^{2}=9+\left(\frac{9}{4}\right)^{2}$$

Square \frac{9}{4} by squaring both the numerator and the denominator of the fraction.

$$k^{2}+\frac{9}{2}k+\frac{81}{16}=9+\frac{81}{16}$$

Add 9 to \frac{81}{16}.

$$k^{2}+\frac{9}{2}k+\frac{81}{16}=\frac{225}{16}$$

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}.

$$\left(k+\frac{9}{4}\right)^{2}=\frac{225}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(k+\frac{9}{4}\right)^{2}}=\sqrt{\frac{225}{16}}$$

Simplify.

$$k+\frac{9}{4}=\frac{15}{4} k+\frac{9}{4}=-\frac{15}{4}$$

Subtract \frac{9}{4} from both sides of the equation.

$$k=\frac{3}{2} k=-6$$