Solve b^2+(17-b)^2=157

$$b ^ { 2 } + ( 17 - b ) ^ { 2 } = 157$$

Solve for b

$$b=6$$

$$b=11$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(17-b\right)^{2}.

$$b^{2}+289-34b+b^{2}=157$$

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

$$2b^{2}+289-34b=157$$

Subtract 157 from both sides.

$$2b^{2}+289-34b-157=0$$

Subtract 157 from 289 to get 132.

$$2b^{2}+132-34b=0$$

Divide both sides by 2.

$$b^{2}+66-17b=0$$

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

$$b^{2}-17b+66=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as b^{2}+ab+bb+66. To find a and b, set up a system to be solved.

$$a+b=-17 ab=1\times 66=66$$

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 66.

$$-1,-66 -2,-33 -3,-22 -6,-11$$

Calculate the sum for each pair.

$$-1-66=-67 -2-33=-35 -3-22=-25 -6-11=-17$$

The solution is the pair that gives sum -17.

$$a=-11 b=-6$$

Rewrite b^{2}-17b+66 as \left(b^{2}-11b\right)+\left(-6b+66\right).

$$\left(b^{2}-11b\right)+\left(-6b+66\right)$$

Factor out b in the first and -6 in the second group.

$$b\left(b-11\right)-6\left(b-11\right)$$

Factor out common term b-11 by using distributive property.

$$\left(b-11\right)\left(b-6\right)$$

To find equation solutions, solve b-11=0 and b-6=0.

$$b=11 b=6$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(17-b\right)^{2}.

$$b^{2}+289-34b+b^{2}=157$$

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

$$2b^{2}+289-34b=157$$

Subtract 157 from both sides.

$$2b^{2}+289-34b-157=0$$

Subtract 157 from 289 to get 132.

$$2b^{2}+132-34b=0$$

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.

$$2b^{2}-34b+132=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -34 for b, and 132 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$b=\frac{-\left(-34\right)±\sqrt{\left(-34\right)^{2}-4\times 2\times 132}}{2\times 2}$$

Square -34.

$$b=\frac{-\left(-34\right)±\sqrt{1156-4\times 2\times 132}}{2\times 2}$$

Multiply -4 times 2.

$$b=\frac{-\left(-34\right)±\sqrt{1156-8\times 132}}{2\times 2}$$

Multiply -8 times 132.

$$b=\frac{-\left(-34\right)±\sqrt{1156-1056}}{2\times 2}$$

Add 1156 to -1056.

$$b=\frac{-\left(-34\right)±\sqrt{100}}{2\times 2}$$

Take the square root of 100.

$$b=\frac{-\left(-34\right)±10}{2\times 2}$$

The opposite of -34 is 34.

$$b=\frac{34±10}{2\times 2}$$

Multiply 2 times 2.

$$b=\frac{34±10}{4}$$

Now solve the equation b=\frac{34±10}{4} when ± is plus. Add 34 to 10.

$$b=\frac{44}{4}$$

Divide 44 by 4.

$$b=11$$

Now solve the equation b=\frac{34±10}{4} when ± is minus. Subtract 10 from 34.

$$b=\frac{24}{4}$$

Divide 24 by 4.

$$b=6$$

The equation is now solved.

$$b=11 b=6$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(17-b\right)^{2}.

$$b^{2}+289-34b+b^{2}=157$$

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

$$2b^{2}+289-34b=157$$

Subtract 289 from both sides.

$$2b^{2}-34b=157-289$$

Subtract 289 from 157 to get -132.

$$2b^{2}-34b=-132$$

Divide both sides by 2.

$$\frac{2b^{2}-34b}{2}=-\frac{132}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$b^{2}+\left(-\frac{34}{2}\right)b=-\frac{132}{2}$$

Divide -34 by 2.

$$b^{2}-17b=-\frac{132}{2}$$

Divide -132 by 2.

$$b^{2}-17b=-66$$

Divide -17, the coefficient of the x term, by 2 to get -\frac{17}{2}. Then add the square of -\frac{17}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$b^{2}-17b+\left(-\frac{17}{2}\right)^{2}=-66+\left(-\frac{17}{2}\right)^{2}$$

Square -\frac{17}{2} by squaring both the numerator and the denominator of the fraction.

$$b^{2}-17b+\frac{289}{4}=-66+\frac{289}{4}$$

Add -66 to \frac{289}{4}.

$$b^{2}-17b+\frac{289}{4}=\frac{25}{4}$$

Factor b^{2}-17b+\frac{289}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(b-\frac{17}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(b-\frac{17}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

$$b-\frac{17}{2}=\frac{5}{2} b-\frac{17}{2}=-\frac{5}{2}$$

Add \frac{17}{2} to both sides of the equation.

$$b=11 b=6$$