Solve 70-15/4h^2=10sqrt{25+h^2/4}

$$70 - \frac { 15 } { 4 } h ^ { 2 } = 10 \sqrt { 25 + \frac { h ^ { 2 } } { 4 } }$$

Solve for h

$$h = \frac{4 \sqrt{11 - \sqrt{67}}}{3} \approx 2.236921388$$

$$h = -\frac{4 \sqrt{11 - \sqrt{67}}}{3} \approx -2.236921388$$

Solution Steps

Square both sides of the equation.

$$\left(70-\frac{15}{4}h^{2}\right)^{2}=\left(10\sqrt{25+\frac{h^{2}}{4}}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(70-\frac{15}{4}h^{2}\right)^{2}.

$$4900-525h^{2}+\frac{225}{16}\left(h^{2}\right)^{2}=\left(10\sqrt{25+\frac{h^{2}}{4}}\right)^{2}$$

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

$$4900-525h^{2}+\frac{225}{16}h^{4}=\left(10\sqrt{25+\frac{h^{2}}{4}}\right)^{2}$$

Expand \left(10\sqrt{25+\frac{h^{2}}{4}}\right)^{2}.

$$4900-525h^{2}+\frac{225}{16}h^{4}=10^{2}\left(\sqrt{25+\frac{h^{2}}{4}}\right)^{2}$$

Calculate 10 to the power of 2 and get 100.

$$4900-525h^{2}+\frac{225}{16}h^{4}=100\left(\sqrt{25+\frac{h^{2}}{4}}\right)^{2}$$

Calculate \sqrt{25+\frac{h^{2}}{4}} to the power of 2 and get 25+\frac{h^{2}}{4}.

$$4900-525h^{2}+\frac{225}{16}h^{4}=100\left(25+\frac{h^{2}}{4}\right)$$

Use the distributive property to multiply 100 by 25+\frac{h^{2}}{4}.

$$4900-525h^{2}+\frac{225}{16}h^{4}=2500+100\times \frac{h^{2}}{4}$$

Cancel out 4, the greatest common factor in 100 and 4.

$$4900-525h^{2}+\frac{225}{16}h^{4}=2500+25h^{2}$$

Subtract 2500 from both sides.

$$4900-525h^{2}+\frac{225}{16}h^{4}-2500=25h^{2}$$

Subtract 2500 from 4900 to get 2400.

$$2400-525h^{2}+\frac{225}{16}h^{4}=25h^{2}$$

Subtract 25h^{2} from both sides.

$$2400-525h^{2}+\frac{225}{16}h^{4}-25h^{2}=0$$

Combine -525h^{2} and -25h^{2} to get -550h^{2}.

$$2400-550h^{2}+\frac{225}{16}h^{4}=0$$

Substitute t for h^{2}.

$$\frac{225}{16}t^{2}-550t+2400=0$$

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{225}{16} for a, -550 for b, and 2400 for c in the quadratic formula.

$$t=\frac{-\left(-550\right)±\sqrt{\left(-550\right)^{2}-4\times \frac{225}{16}\times 2400}}{2\times \frac{225}{16}}$$

Do the calculations.

$$t=\frac{550±50\sqrt{67}}{\frac{225}{8}}$$

Solve the equation t=\frac{550±50\sqrt{67}}{\frac{225}{8}} when ± is plus and when ± is minus.

$$t=\frac{16\sqrt{67}+176}{9} t=\frac{176-16\sqrt{67}}{9}$$

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

$$h=\frac{4\sqrt{\sqrt{67}+11}}{3} h=-\frac{4\sqrt{\sqrt{67}+11}}{3} h=\frac{4\sqrt{11-\sqrt{67}}}{3} h=-\frac{4\sqrt{11-\sqrt{67}}}{3}$$

Substitute \frac{4\sqrt{\sqrt{67}+11}}{3} for h in the equation 70-\frac{15}{4}h^{2}=10\sqrt{25+\frac{h^{2}}{4}}.

$$70-\frac{15}{4}\times \left(\frac{4\sqrt{\sqrt{67}+11}}{3}\right)^{2}=10\sqrt{25+\frac{\left(\frac{4\sqrt{\sqrt{67}+11}}{3}\right)^{2}}{4}}$$

Simplify. The value h=\frac{4\sqrt{\sqrt{67}+11}}{3} does not satisfy the equation because the left and the right hand side have opposite signs.

$$-\frac{10}{3}-\frac{20}{3}\times 67^{\frac{1}{2}}=\frac{20}{3}\times 67^{\frac{1}{2}}+\frac{10}{3}$$

Substitute -\frac{4\sqrt{\sqrt{67}+11}}{3} for h in the equation 70-\frac{15}{4}h^{2}=10\sqrt{25+\frac{h^{2}}{4}}.

$$70-\frac{15}{4}\left(-\frac{4\sqrt{\sqrt{67}+11}}{3}\right)^{2}=10\sqrt{25+\frac{\left(-\frac{4\sqrt{\sqrt{67}+11}}{3}\right)^{2}}{4}}$$

Simplify. The value h=-\frac{4\sqrt{\sqrt{67}+11}}{3} does not satisfy the equation because the left and the right hand side have opposite signs.

$$-\frac{10}{3}-\frac{20}{3}\times 67^{\frac{1}{2}}=\frac{20}{3}\times 67^{\frac{1}{2}}+\frac{10}{3}$$

Substitute \frac{4\sqrt{11-\sqrt{67}}}{3} for h in the equation 70-\frac{15}{4}h^{2}=10\sqrt{25+\frac{h^{2}}{4}}.

$$70-\frac{15}{4}\times \left(\frac{4\sqrt{11-\sqrt{67}}}{3}\right)^{2}=10\sqrt{25+\frac{\left(\frac{4\sqrt{11-\sqrt{67}}}{3}\right)^{2}}{4}}$$

Simplify. The value h=\frac{4\sqrt{11-\sqrt{67}}}{3} satisfies the equation.

$$-\frac{10}{3}+\frac{20}{3}\times 67^{\frac{1}{2}}=\frac{20}{3}\times 67^{\frac{1}{2}}-\frac{10}{3}$$

Substitute -\frac{4\sqrt{11-\sqrt{67}}}{3} for h in the equation 70-\frac{15}{4}h^{2}=10\sqrt{25+\frac{h^{2}}{4}}.

$$70-\frac{15}{4}\left(-\frac{4\sqrt{11-\sqrt{67}}}{3}\right)^{2}=10\sqrt{25+\frac{\left(-\frac{4\sqrt{11-\sqrt{67}}}{3}\right)^{2}}{4}}$$

Simplify. The value h=-\frac{4\sqrt{11-\sqrt{67}}}{3} satisfies the equation.

$$-\frac{10}{3}+\frac{20}{3}\times 67^{\frac{1}{2}}=\frac{20}{3}\times 67^{\frac{1}{2}}-\frac{10}{3}$$

List all solutions of -\frac{15h^{2}}{4}+70=10\sqrt{\frac{h^{2}}{4}+25}.

$$h=\frac{4\sqrt{11-\sqrt{67}}}{3} h=-\frac{4\sqrt{11-\sqrt{67}}}{3}$$