Solve (8-k)^2-(2k+2)^2/(8-k)^2=0

$$\frac { ( 8 - k ) ^ { 2 } - ( 2 k + 2 ) ^ { 2 } } { ( 8 - k ) ^ { 2 } } = 0$$

Solve for k

$$k=-10$$

$$k=2$$

Steps Using Factoring By Grouping

Variable k cannot be equal to 8 since division by zero is not defined. Multiply both sides of the equation by \left(k-8\right)^{2}.

$$\left(8-k\right)^{2}-\left(2k+2\right)^{2}=0$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(8-k\right)^{2}.

$$64-16k+k^{2}-\left(2k+2\right)^{2}=0$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2k+2\right)^{2}.

$$64-16k+k^{2}-\left(4k^{2}+8k+4\right)=0$$

To find the opposite of 4k^{2}+8k+4, find the opposite of each term.

$$64-16k+k^{2}-4k^{2}-8k-4=0$$

Combine k^{2} and -4k^{2} to get -3k^{2}.

$$64-16k-3k^{2}-8k-4=0$$

Combine -16k and -8k to get -24k.

$$64-24k-3k^{2}-4=0$$

Subtract 4 from 64 to get 60.

$$60-24k-3k^{2}=0$$

Divide both sides by 3.

$$20-8k-k^{2}=0$$

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

$$-k^{2}-8k+20=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -k^{2}+ak+bk+20. To find a and b, set up a system to be solved.

$$a+b=-8 ab=-20=-20$$

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 -20.

$$1,-20 2,-10 4,-5$$

Calculate the sum for each pair.

$$1-20=-19 2-10=-8 4-5=-1$$

The solution is the pair that gives sum -8.

$$a=2 b=-10$$

Rewrite -k^{2}-8k+20 as \left(-k^{2}+2k\right)+\left(-10k+20\right).

$$\left(-k^{2}+2k\right)+\left(-10k+20\right)$$

Factor out k in the first and 10 in the second group.

$$k\left(-k+2\right)+10\left(-k+2\right)$$

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

$$\left(-k+2\right)\left(k+10\right)$$

To find equation solutions, solve -k+2=0 and k+10=0.

$$k=2 k=-10$$

Steps Using the Quadratic Formula

Variable k cannot be equal to 8 since division by zero is not defined. Multiply both sides of the equation by \left(k-8\right)^{2}.

$$\left(8-k\right)^{2}-\left(2k+2\right)^{2}=0$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(8-k\right)^{2}.

$$64-16k+k^{2}-\left(2k+2\right)^{2}=0$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2k+2\right)^{2}.

$$64-16k+k^{2}-\left(4k^{2}+8k+4\right)=0$$

To find the opposite of 4k^{2}+8k+4, find the opposite of each term.

$$64-16k+k^{2}-4k^{2}-8k-4=0$$

Combine k^{2} and -4k^{2} to get -3k^{2}.

$$64-16k-3k^{2}-8k-4=0$$

Combine -16k and -8k to get -24k.

$$64-24k-3k^{2}-4=0$$

Subtract 4 from 64 to get 60.

$$60-24k-3k^{2}=0$$

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.

$$-3k^{2}-24k+60=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, -24 for b, and 60 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$k=\frac{-\left(-24\right)±\sqrt{\left(-24\right)^{2}-4\left(-3\right)\times 60}}{2\left(-3\right)}$$

Square -24.

$$k=\frac{-\left(-24\right)±\sqrt{576-4\left(-3\right)\times 60}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$k=\frac{-\left(-24\right)±\sqrt{576+12\times 60}}{2\left(-3\right)}$$

Multiply 12 times 60.

$$k=\frac{-\left(-24\right)±\sqrt{576+720}}{2\left(-3\right)}$$

Add 576 to 720.

$$k=\frac{-\left(-24\right)±\sqrt{1296}}{2\left(-3\right)}$$

Take the square root of 1296.

$$k=\frac{-\left(-24\right)±36}{2\left(-3\right)}$$

The opposite of -24 is 24.

$$k=\frac{24±36}{2\left(-3\right)}$$

Multiply 2 times -3.

$$k=\frac{24±36}{-6}$$

Now solve the equation k=\frac{24±36}{-6} when ± is plus. Add 24 to 36.

$$k=\frac{60}{-6}$$

Divide 60 by -6.

$$k=-10$$

Now solve the equation k=\frac{24±36}{-6} when ± is minus. Subtract 36 from 24.

$$k=-\frac{12}{-6}$$

Divide -12 by -6.

$$k=2$$

The equation is now solved.

$$k=-10 k=2$$

Steps for Completing the Square

Variable k cannot be equal to 8 since division by zero is not defined. Multiply both sides of the equation by \left(k-8\right)^{2}.

$$\left(8-k\right)^{2}-\left(2k+2\right)^{2}=0$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(8-k\right)^{2}.

$$64-16k+k^{2}-\left(2k+2\right)^{2}=0$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2k+2\right)^{2}.

$$64-16k+k^{2}-\left(4k^{2}+8k+4\right)=0$$

To find the opposite of 4k^{2}+8k+4, find the opposite of each term.

$$64-16k+k^{2}-4k^{2}-8k-4=0$$

Combine k^{2} and -4k^{2} to get -3k^{2}.

$$64-16k-3k^{2}-8k-4=0$$

Combine -16k and -8k to get -24k.

$$64-24k-3k^{2}-4=0$$

Subtract 4 from 64 to get 60.

$$60-24k-3k^{2}=0$$

Subtract 60 from both sides. Anything subtracted from zero gives its negation.

$$-24k-3k^{2}=-60$$

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.

$$-3k^{2}-24k=-60$$

Divide both sides by -3.

$$\frac{-3k^{2}-24k}{-3}=-\frac{60}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$k^{2}+\left(-\frac{24}{-3}\right)k=-\frac{60}{-3}$$

Divide -24 by -3.

$$k^{2}+8k=-\frac{60}{-3}$$

Divide -60 by -3.

$$k^{2}+8k=20$$

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

$$k^{2}+8k+4^{2}=20+4^{2}$$

Square 4.

$$k^{2}+8k+16=20+16$$

Add 20 to 16.

$$k^{2}+8k+16=36$$

Factor k^{2}+8k+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}.

$$\left(k+4\right)^{2}=36$$

Take the square root of both sides of the equation.

$$\sqrt{\left(k+4\right)^{2}}=\sqrt{36}$$

Simplify.

$$k+4=6 k+4=-6$$

Subtract 4 from both sides of the equation.

$$k=2 k=-10$$