Solve 8-3/2b=sqrt{25-3/4b^2}

$$8 - \frac { 3 } { 2 } b = \sqrt { 25 - \frac { 3 } { 4 } b ^ { 2 } }$$

Solve for b

$$b=4-\sqrt{3}\approx 2.267949192$$

Solution Steps

Square both sides of the equation.

$$\left(8-\frac{3}{2}b\right)^{2}=\left(\sqrt{25-\frac{3}{4}b^{2}}\right)^{2}$$

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

$$64-24b+\frac{9}{4}b^{2}=\left(\sqrt{25-\frac{3}{4}b^{2}}\right)^{2}$$

Calculate \sqrt{25-\frac{3}{4}b^{2}} to the power of 2 and get 25-\frac{3}{4}b^{2}.

$$64-24b+\frac{9}{4}b^{2}=25-\frac{3}{4}b^{2}$$

Subtract 25 from both sides.

$$64-24b+\frac{9}{4}b^{2}-25=-\frac{3}{4}b^{2}$$

Subtract 25 from 64 to get 39.

$$39-24b+\frac{9}{4}b^{2}=-\frac{3}{4}b^{2}$$

Add \frac{3}{4}b^{2} to both sides.

$$39-24b+\frac{9}{4}b^{2}+\frac{3}{4}b^{2}=0$$

Combine \frac{9}{4}b^{2} and \frac{3}{4}b^{2} to get 3b^{2}.

$$39-24b+3b^{2}=0$$

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.

$$3b^{2}-24b+39=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -24 for b, and 39 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$b=\frac{-\left(-24\right)±\sqrt{\left(-24\right)^{2}-4\times 3\times 39}}{2\times 3}$$

Square -24.

$$b=\frac{-\left(-24\right)±\sqrt{576-4\times 3\times 39}}{2\times 3}$$

Multiply -4 times 3.

$$b=\frac{-\left(-24\right)±\sqrt{576-12\times 39}}{2\times 3}$$

Multiply -12 times 39.

$$b=\frac{-\left(-24\right)±\sqrt{576-468}}{2\times 3}$$

Add 576 to -468.

$$b=\frac{-\left(-24\right)±\sqrt{108}}{2\times 3}$$

Take the square root of 108.

$$b=\frac{-\left(-24\right)±6\sqrt{3}}{2\times 3}$$

The opposite of -24 is 24.

$$b=\frac{24±6\sqrt{3}}{2\times 3}$$

Multiply 2 times 3.

$$b=\frac{24±6\sqrt{3}}{6}$$

Now solve the equation b=\frac{24±6\sqrt{3}}{6} when ± is plus. Add 24 to 6\sqrt{3}.

$$b=\frac{6\sqrt{3}+24}{6}$$

Divide 24+6\sqrt{3} by 6.

$$b=\sqrt{3}+4$$

Now solve the equation b=\frac{24±6\sqrt{3}}{6} when ± is minus. Subtract 6\sqrt{3} from 24.

$$b=\frac{24-6\sqrt{3}}{6}$$

Divide 24-6\sqrt{3} by 6.

$$b=4-\sqrt{3}$$

The equation is now solved.

$$b=\sqrt{3}+4 b=4-\sqrt{3}$$

Substitute \sqrt{3}+4 for b in the equation 8-\frac{3}{2}b=\sqrt{25-\frac{3}{4}b^{2}}.

$$8-\frac{3}{2}\left(\sqrt{3}+4\right)=\sqrt{25-\frac{3}{4}\left(\sqrt{3}+4\right)^{2}}$$

Simplify. The value b=\sqrt{3}+4 does not satisfy the equation because the left and the right hand side have opposite signs.

$$2-\frac{3}{2}\times 3^{\frac{1}{2}}=\frac{3}{2}\times 3^{\frac{1}{2}}-2$$

Substitute 4-\sqrt{3} for b in the equation 8-\frac{3}{2}b=\sqrt{25-\frac{3}{4}b^{2}}.

$$8-\frac{3}{2}\left(4-\sqrt{3}\right)=\sqrt{25-\frac{3}{4}\left(4-\sqrt{3}\right)^{2}}$$

Simplify. The value b=4-\sqrt{3} satisfies the equation.

$$2+\frac{3}{2}\times 3^{\frac{1}{2}}=\frac{3}{2}\times 3^{\frac{1}{2}}+2$$

Equation -\frac{3b}{2}+8=\sqrt{-\frac{3b^{2}}{4}+25} has a unique solution.

$$b=4-\sqrt{3}$$