Solve (15-b^2/sqrt{2})^2+1=9

$$( \frac { 15 - b ^ { 2 } } { \sqrt { 2 } } ) ^ { 2 } + 1 = 9$$

Solve for b

$$b=-\sqrt{19}\approx -4.358898944$$

$$b=\sqrt{19}\approx 4.358898944$$

$$b=\sqrt{11}\approx 3.31662479$$

$$b=-\sqrt{11}\approx -3.31662479$$

Steps Using the Quadratic Formula

Rationalize the denominator of \frac{15-b^{2}}{\sqrt{2}} by multiplying numerator and denominator by \sqrt{2}.

$$\left(\frac{\left(15-b^{2}\right)\sqrt{2}}{\left(\sqrt{2}\right)^{2}}\right)^{2}+1=9$$

The square of \sqrt{2} is 2.

$$\left(\frac{\left(15-b^{2}\right)\sqrt{2}}{2}\right)^{2}+1=9$$

To raise \frac{\left(15-b^{2}\right)\sqrt{2}}{2} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(\left(15-b^{2}\right)\sqrt{2}\right)^{2}}{2^{2}}+1=9$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{2^{2}}{2^{2}}.

$$\frac{\left(\left(15-b^{2}\right)\sqrt{2}\right)^{2}}{2^{2}}+\frac{2^{2}}{2^{2}}=9$$

Since \frac{\left(\left(15-b^{2}\right)\sqrt{2}\right)^{2}}{2^{2}} and \frac{2^{2}}{2^{2}} have the same denominator, add them by adding their numerators.

$$\frac{\left(\left(15-b^{2}\right)\sqrt{2}\right)^{2}+2^{2}}{2^{2}}=9$$

Expand \left(\left(15-b^{2}\right)\sqrt{2}\right)^{2}.

$$\frac{\left(15-b^{2}\right)^{2}\left(\sqrt{2}\right)^{2}+2^{2}}{2^{2}}=9$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(15-b^{2}\right)^{2}.

$$\frac{\left(225-30b^{2}+\left(b^{2}\right)^{2}\right)\left(\sqrt{2}\right)^{2}+2^{2}}{2^{2}}=9$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$\frac{\left(225-30b^{2}+b^{4}\right)\left(\sqrt{2}\right)^{2}+2^{2}}{2^{2}}=9$$

The square of \sqrt{2} is 2.

$$\frac{\left(225-30b^{2}+b^{4}\right)\times 2+2^{2}}{2^{2}}=9$$

Calculate 2 to the power of 2 and get 4.

$$\frac{\left(225-30b^{2}+b^{4}\right)\times 2+4}{2^{2}}=9$$

Calculate 2 to the power of 2 and get 4.

$$\frac{\left(225-30b^{2}+b^{4}\right)\times 2+4}{4}=9$$

Divide each term of \left(225-30b^{2}+b^{4}\right)\times 2+4 by 4 to get \frac{1}{2}\left(225-30b^{2}+b^{4}\right)+1.

$$\frac{1}{2}\left(225-30b^{2}+b^{4}\right)+1=9$$

Use the distributive property to multiply \frac{1}{2} by 225-30b^{2}+b^{4}.

$$\frac{225}{2}-15b^{2}+\frac{1}{2}b^{4}+1=9$$

Add \frac{225}{2} and 1 to get \frac{227}{2}.

$$\frac{227}{2}-15b^{2}+\frac{1}{2}b^{4}=9$$

Subtract 9 from both sides.

$$\frac{227}{2}-15b^{2}+\frac{1}{2}b^{4}-9=0$$

Subtract 9 from \frac{227}{2} to get \frac{209}{2}.

$$\frac{209}{2}-15b^{2}+\frac{1}{2}b^{4}=0$$

Substitute t for b^{2}.

$$\frac{1}{2}t^{2}-15t+\frac{209}{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute \frac{1}{2} for a, -15 for b, and \frac{209}{2} for c in the quadratic formula.

$$t=\frac{-\left(-15\right)±\sqrt{\left(-15\right)^{2}-4\times \frac{1}{2}\times \frac{209}{2}}}{\frac{1}{2}\times 2}$$

Do the calculations.

$$t=\frac{15±4}{1}$$

Solve the equation t=\frac{15±4}{1} when ± is plus and when ± is minus.

$$t=19 t=11$$

Since b=t^{2}, the solutions are obtained by evaluating b=±\sqrt{t} for each t.

$$b=\sqrt{19} b=-\sqrt{19} b=\sqrt{11} b=-\sqrt{11}$$