Solve (5-5k)^2+9=9k^2

$$( 5 - 5 k ) ^ { 2 } + 9 = 9 k ^ { 2 }$$

Solve for k

$$k=1$$

$$k = \frac{17}{8} = 2\frac{1}{8} = 2.125$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5-5k\right)^{2}.

$$25-50k+25k^{2}+9=9k^{2}$$

Add 25 and 9 to get 34.

$$34-50k+25k^{2}=9k^{2}$$

Subtract 9k^{2} from both sides.

$$34-50k+25k^{2}-9k^{2}=0$$

Combine 25k^{2} and -9k^{2} to get 16k^{2}.

$$34-50k+16k^{2}=0$$

Divide both sides by 2.

$$17-25k+8k^{2}=0$$

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

$$8k^{2}-25k+17=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 8k^{2}+ak+bk+17. To find a and b, set up a system to be solved.

$$a+b=-25 ab=8\times 17=136$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 136.

$$-1,-136 -2,-68 -4,-34 -8,-17$$

Calculate the sum for each pair.

$$-1-136=-137 -2-68=-70 -4-34=-38 -8-17=-25$$

The solution is the pair that gives sum -25.

$$a=-17 b=-8$$

Rewrite 8k^{2}-25k+17 as \left(8k^{2}-17k\right)+\left(-8k+17\right).

$$\left(8k^{2}-17k\right)+\left(-8k+17\right)$$

Factor out k in the first and -1 in the second group.

$$k\left(8k-17\right)-\left(8k-17\right)$$

Factor out common term 8k-17 by using distributive property.

$$\left(8k-17\right)\left(k-1\right)$$

To find equation solutions, solve 8k-17=0 and k-1=0.

$$k=\frac{17}{8} k=1$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5-5k\right)^{2}.

$$25-50k+25k^{2}+9=9k^{2}$$

Add 25 and 9 to get 34.

$$34-50k+25k^{2}=9k^{2}$$

Subtract 9k^{2} from both sides.

$$34-50k+25k^{2}-9k^{2}=0$$

Combine 25k^{2} and -9k^{2} to get 16k^{2}.

$$34-50k+16k^{2}=0$$

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.

$$16k^{2}-50k+34=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 16 for a, -50 for b, and 34 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$k=\frac{-\left(-50\right)±\sqrt{\left(-50\right)^{2}-4\times 16\times 34}}{2\times 16}$$

Square -50.

$$k=\frac{-\left(-50\right)±\sqrt{2500-4\times 16\times 34}}{2\times 16}$$

Multiply -4 times 16.

$$k=\frac{-\left(-50\right)±\sqrt{2500-64\times 34}}{2\times 16}$$

Multiply -64 times 34.

$$k=\frac{-\left(-50\right)±\sqrt{2500-2176}}{2\times 16}$$

Add 2500 to -2176.

$$k=\frac{-\left(-50\right)±\sqrt{324}}{2\times 16}$$

Take the square root of 324.

$$k=\frac{-\left(-50\right)±18}{2\times 16}$$

The opposite of -50 is 50.

$$k=\frac{50±18}{2\times 16}$$

Multiply 2 times 16.

$$k=\frac{50±18}{32}$$

Now solve the equation k=\frac{50±18}{32} when ± is plus. Add 50 to 18.

$$k=\frac{68}{32}$$

Reduce the fraction \frac{68}{32} to lowest terms by extracting and canceling out 4.

$$k=\frac{17}{8}$$

Now solve the equation k=\frac{50±18}{32} when ± is minus. Subtract 18 from 50.

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

Divide 32 by 32.

$$k=1$$

The equation is now solved.

$$k=\frac{17}{8} k=1$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5-5k\right)^{2}.

$$25-50k+25k^{2}+9=9k^{2}$$

Add 25 and 9 to get 34.

$$34-50k+25k^{2}=9k^{2}$$

Subtract 9k^{2} from both sides.

$$34-50k+25k^{2}-9k^{2}=0$$

Combine 25k^{2} and -9k^{2} to get 16k^{2}.

$$34-50k+16k^{2}=0$$

Subtract 34 from both sides. Anything subtracted from zero gives its negation.

$$-50k+16k^{2}=-34$$

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.

$$16k^{2}-50k=-34$$

Divide both sides by 16.

$$\frac{16k^{2}-50k}{16}=-\frac{34}{16}$$

Dividing by 16 undoes the multiplication by 16.

$$k^{2}+\left(-\frac{50}{16}\right)k=-\frac{34}{16}$$

Reduce the fraction \frac{-50}{16} to lowest terms by extracting and canceling out 2.

$$k^{2}-\frac{25}{8}k=-\frac{34}{16}$$

Reduce the fraction \frac{-34}{16} to lowest terms by extracting and canceling out 2.

$$k^{2}-\frac{25}{8}k=-\frac{17}{8}$$

Divide -\frac{25}{8}, the coefficient of the x term, by 2 to get -\frac{25}{16}. Then add the square of -\frac{25}{16} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$k^{2}-\frac{25}{8}k+\left(-\frac{25}{16}\right)^{2}=-\frac{17}{8}+\left(-\frac{25}{16}\right)^{2}$$

Square -\frac{25}{16} by squaring both the numerator and the denominator of the fraction.

$$k^{2}-\frac{25}{8}k+\frac{625}{256}=-\frac{17}{8}+\frac{625}{256}$$

Add -\frac{17}{8} to \frac{625}{256} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$k^{2}-\frac{25}{8}k+\frac{625}{256}=\frac{81}{256}$$

Factor k^{2}-\frac{25}{8}k+\frac{625}{256}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(k-\frac{25}{16}\right)^{2}=\frac{81}{256}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(k-\frac{25}{16}\right)^{2}}=\sqrt{\frac{81}{256}}$$

Simplify.

$$k-\frac{25}{16}=\frac{9}{16} k-\frac{25}{16}=-\frac{9}{16}$$

Add \frac{25}{16} to both sides of the equation.

$$k=\frac{17}{8} k=1$$