Solve 0=(-4k)^2-4*(k+1)*9

$$0 = ( - 4 k ) ^ { 2 } - 4 \times ( k + 1 ) \times 9$$

Solve for k

$$k=-\frac{3}{4}=-0.75$$

$$k=3$$

Steps Using Factoring By Grouping

Expand \left(-4k\right)^{2}.

$$0=\left(-4\right)^{2}k^{2}-4\left(k+1\right)\times 9$$

Calculate -4 to the power of 2 and get 16.

$$0=16k^{2}-4\left(k+1\right)\times 9$$

Multiply 4 and 9 to get 36.

$$0=16k^{2}-36\left(k+1\right)$$

Use the distributive property to multiply -36 by k+1.

$$0=16k^{2}-36k-36$$

Swap sides so that all variable terms are on the left hand side.

$$16k^{2}-36k-36=0$$

Divide both sides by 4.

$$4k^{2}-9k-9=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4k^{2}+ak+bk-9. To find a and b, set up a system to be solved.

$$a+b=-9 ab=4\left(-9\right)=-36$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. 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.

$$a=-12 b=3$$

Rewrite 4k^{2}-9k-9 as \left(4k^{2}-12k\right)+\left(3k-9\right).

$$\left(4k^{2}-12k\right)+\left(3k-9\right)$$

Factor out 4k in the first and 3 in the second group.

$$4k\left(k-3\right)+3\left(k-3\right)$$

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

$$\left(k-3\right)\left(4k+3\right)$$

To find equation solutions, solve k-3=0 and 4k+3=0.

$$k=3 k=-\frac{3}{4}$$

Steps Using the Quadratic Formula

Expand \left(-4k\right)^{2}.

$$0=\left(-4\right)^{2}k^{2}-4\left(k+1\right)\times 9$$

Calculate -4 to the power of 2 and get 16.

$$0=16k^{2}-4\left(k+1\right)\times 9$$

Multiply 4 and 9 to get 36.

$$0=16k^{2}-36\left(k+1\right)$$

Use the distributive property to multiply -36 by k+1.

$$0=16k^{2}-36k-36$$

Swap sides so that all variable terms are on the left hand side.

$$16k^{2}-36k-36=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 16 for a, -36 for b, and -36 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$k=\frac{-\left(-36\right)±\sqrt{\left(-36\right)^{2}-4\times 16\left(-36\right)}}{2\times 16}$$

Square -36.

$$k=\frac{-\left(-36\right)±\sqrt{1296-4\times 16\left(-36\right)}}{2\times 16}$$

Multiply -4 times 16.

$$k=\frac{-\left(-36\right)±\sqrt{1296-64\left(-36\right)}}{2\times 16}$$

Multiply -64 times -36.

$$k=\frac{-\left(-36\right)±\sqrt{1296+2304}}{2\times 16}$$

Add 1296 to 2304.

$$k=\frac{-\left(-36\right)±\sqrt{3600}}{2\times 16}$$

Take the square root of 3600.

$$k=\frac{-\left(-36\right)±60}{2\times 16}$$

The opposite of -36 is 36.

$$k=\frac{36±60}{2\times 16}$$

Multiply 2 times 16.

$$k=\frac{36±60}{32}$$

Now solve the equation k=\frac{36±60}{32} when ± is plus. Add 36 to 60.

$$k=\frac{96}{32}$$

Divide 96 by 32.

$$k=3$$

Now solve the equation k=\frac{36±60}{32} when ± is minus. Subtract 60 from 36.

$$k=-\frac{24}{32}$$

Reduce the fraction \frac{-24}{32} to lowest terms by extracting and canceling out 8.

$$k=-\frac{3}{4}$$

The equation is now solved.

$$k=3 k=-\frac{3}{4}$$

Steps for Completing the Square

Expand \left(-4k\right)^{2}.

$$0=\left(-4\right)^{2}k^{2}-4\left(k+1\right)\times 9$$

Calculate -4 to the power of 2 and get 16.

$$0=16k^{2}-4\left(k+1\right)\times 9$$

Multiply 4 and 9 to get 36.

$$0=16k^{2}-36\left(k+1\right)$$

Use the distributive property to multiply -36 by k+1.

$$0=16k^{2}-36k-36$$

Swap sides so that all variable terms are on the left hand side.

$$16k^{2}-36k-36=0$$

Add 36 to both sides. Anything plus zero gives itself.

$$16k^{2}-36k=36$$

Divide both sides by 16.

$$\frac{16k^{2}-36k}{16}=\frac{36}{16}$$

Dividing by 16 undoes the multiplication by 16.

$$k^{2}+\left(-\frac{36}{16}\right)k=\frac{36}{16}$$

Reduce the fraction \frac{-36}{16} to lowest terms by extracting and canceling out 4.

$$k^{2}-\frac{9}{4}k=\frac{36}{16}$$

Reduce the fraction \frac{36}{16} to lowest terms by extracting and canceling out 4.

$$k^{2}-\frac{9}{4}k=\frac{9}{4}$$

Divide -\frac{9}{4}, the coefficient of the x term, by 2 to get -\frac{9}{8}. Then add the square of -\frac{9}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$k^{2}-\frac{9}{4}k+\left(-\frac{9}{8}\right)^{2}=\frac{9}{4}+\left(-\frac{9}{8}\right)^{2}$$

Square -\frac{9}{8} by squaring both the numerator and the denominator of the fraction.

$$k^{2}-\frac{9}{4}k+\frac{81}{64}=\frac{9}{4}+\frac{81}{64}$$

Add \frac{9}{4} to \frac{81}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$k^{2}-\frac{9}{4}k+\frac{81}{64}=\frac{225}{64}$$

Factor k^{2}-\frac{9}{4}k+\frac{81}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(k-\frac{9}{8}\right)^{2}=\frac{225}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(k-\frac{9}{8}\right)^{2}}=\sqrt{\frac{225}{64}}$$

Simplify.

$$k-\frac{9}{8}=\frac{15}{8} k-\frac{9}{8}=-\frac{15}{8}$$

Add \frac{9}{8} to both sides of the equation.

$$k=3 k=-\frac{3}{4}$$