Solve 8*(3b-8)^2-(9-b)^2=0

$$8 \cdot ( 3 b - 8 ) ^ { 2 } - ( 9 - b ) ^ { 2 } = 0$$

Solve for b

$$b = \frac{38 \sqrt{2} + 183}{71} \approx 3.334367822$$

$$b = \frac{183 - 38 \sqrt{2}}{71} \approx 1.820561755$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3b-8\right)^{2}.

$$8\left(9b^{2}-48b+64\right)-\left(9-b\right)^{2}=0$$

Use the distributive property to multiply 8 by 9b^{2}-48b+64.

$$72b^{2}-384b+512-\left(9-b\right)^{2}=0$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(9-b\right)^{2}.

$$72b^{2}-384b+512-\left(81-18b+b^{2}\right)=0$$

To find the opposite of 81-18b+b^{2}, find the opposite of each term.

$$72b^{2}-384b+512-81+18b-b^{2}=0$$

Subtract 81 from 512 to get 431.

$$72b^{2}-384b+431+18b-b^{2}=0$$

Combine -384b and 18b to get -366b.

$$72b^{2}-366b+431-b^{2}=0$$

Combine 72b^{2} and -b^{2} to get 71b^{2}.

$$71b^{2}-366b+431=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 71 for a, -366 for b, and 431 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$b=\frac{-\left(-366\right)±\sqrt{\left(-366\right)^{2}-4\times 71\times 431}}{2\times 71}$$

Square -366.

$$b=\frac{-\left(-366\right)±\sqrt{133956-4\times 71\times 431}}{2\times 71}$$

Multiply -4 times 71.

$$b=\frac{-\left(-366\right)±\sqrt{133956-284\times 431}}{2\times 71}$$

Multiply -284 times 431.

$$b=\frac{-\left(-366\right)±\sqrt{133956-122404}}{2\times 71}$$

Add 133956 to -122404.

$$b=\frac{-\left(-366\right)±\sqrt{11552}}{2\times 71}$$

Take the square root of 11552.

$$b=\frac{-\left(-366\right)±76\sqrt{2}}{2\times 71}$$

The opposite of -366 is 366.

$$b=\frac{366±76\sqrt{2}}{2\times 71}$$

Multiply 2 times 71.

$$b=\frac{366±76\sqrt{2}}{142}$$

Now solve the equation b=\frac{366±76\sqrt{2}}{142} when ± is plus. Add 366 to 76\sqrt{2}.

$$b=\frac{76\sqrt{2}+366}{142}$$

Divide 366+76\sqrt{2} by 142.

$$b=\frac{38\sqrt{2}+183}{71}$$

Now solve the equation b=\frac{366±76\sqrt{2}}{142} when ± is minus. Subtract 76\sqrt{2} from 366.

$$b=\frac{366-76\sqrt{2}}{142}$$

Divide 366-76\sqrt{2} by 142.

$$b=\frac{183-38\sqrt{2}}{71}$$

The equation is now solved.

$$b=\frac{38\sqrt{2}+183}{71} b=\frac{183-38\sqrt{2}}{71}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3b-8\right)^{2}.

$$8\left(9b^{2}-48b+64\right)-\left(9-b\right)^{2}=0$$

Use the distributive property to multiply 8 by 9b^{2}-48b+64.

$$72b^{2}-384b+512-\left(9-b\right)^{2}=0$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(9-b\right)^{2}.

$$72b^{2}-384b+512-\left(81-18b+b^{2}\right)=0$$

To find the opposite of 81-18b+b^{2}, find the opposite of each term.

$$72b^{2}-384b+512-81+18b-b^{2}=0$$

Subtract 81 from 512 to get 431.

$$72b^{2}-384b+431+18b-b^{2}=0$$

Combine -384b and 18b to get -366b.

$$72b^{2}-366b+431-b^{2}=0$$

Combine 72b^{2} and -b^{2} to get 71b^{2}.

$$71b^{2}-366b+431=0$$

Subtract 431 from both sides. Anything subtracted from zero gives its negation.

$$71b^{2}-366b=-431$$

Divide both sides by 71.

$$\frac{71b^{2}-366b}{71}=-\frac{431}{71}$$

Dividing by 71 undoes the multiplication by 71.

$$b^{2}-\frac{366}{71}b=-\frac{431}{71}$$

Divide -\frac{366}{71}, the coefficient of the x term, by 2 to get -\frac{183}{71}. Then add the square of -\frac{183}{71} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$b^{2}-\frac{366}{71}b+\left(-\frac{183}{71}\right)^{2}=-\frac{431}{71}+\left(-\frac{183}{71}\right)^{2}$$

Square -\frac{183}{71} by squaring both the numerator and the denominator of the fraction.

$$b^{2}-\frac{366}{71}b+\frac{33489}{5041}=-\frac{431}{71}+\frac{33489}{5041}$$

Add -\frac{431}{71} to \frac{33489}{5041} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$b^{2}-\frac{366}{71}b+\frac{33489}{5041}=\frac{2888}{5041}$$

Factor b^{2}-\frac{366}{71}b+\frac{33489}{5041}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(b-\frac{183}{71}\right)^{2}=\frac{2888}{5041}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(b-\frac{183}{71}\right)^{2}}=\sqrt{\frac{2888}{5041}}$$

Simplify.

$$b-\frac{183}{71}=\frac{38\sqrt{2}}{71} b-\frac{183}{71}=-\frac{38\sqrt{2}}{71}$$

Add \frac{183}{71} to both sides of the equation.

$$b=\frac{38\sqrt{2}+183}{71} b=\frac{183-38\sqrt{2}}{71}$$