Solve sqrt{(8-t-3/5t)^2+(4/5)^2}=t

$$\sqrt { ( 8 - t - \frac { 3 } { 5 } t ) ^ { 2 } + ( \frac { 4 } { 5 } ) ^ { 2 } } = t$$

Solve for t

$$t = \frac{4 \sqrt{2461} + 320}{39} \approx 13.293176105$$

$$t = \frac{320 - 4 \sqrt{2461}}{39} \approx 3.117080306$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{\left(8-t-\frac{3}{5}t\right)^{2}+\left(\frac{4}{5}\right)^{2}}\right)^{2}=t^{2}$$

Combine -t and -\frac{3}{5}t to get -\frac{8}{5}t.

$$\left(\sqrt{\left(8-\frac{8}{5}t\right)^{2}+\left(\frac{4}{5}\right)^{2}}\right)^{2}=t^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(8-\frac{8}{5}t\right)^{2}.

$$\left(\sqrt{64-\frac{128}{5}t+\frac{64}{25}t^{2}+\left(\frac{4}{5}\right)^{2}}\right)^{2}=t^{2}$$

Calculate \frac{4}{5} to the power of 2 and get \frac{16}{25}.

$$\left(\sqrt{64-\frac{128}{5}t+\frac{64}{25}t^{2}+\frac{16}{25}}\right)^{2}=t^{2}$$

Add 64 and \frac{16}{25} to get \frac{1616}{25}.

$$\left(\sqrt{\frac{1616}{25}-\frac{128}{5}t+\frac{64}{25}t^{2}}\right)^{2}=t^{2}$$

Calculate \sqrt{\frac{1616}{25}-\frac{128}{5}t+\frac{64}{25}t^{2}} to the power of 2 and get \frac{1616}{25}-\frac{128}{5}t+\frac{64}{25}t^{2}.

$$\frac{1616}{25}-\frac{128}{5}t+\frac{64}{25}t^{2}=t^{2}$$

Subtract t^{2} from both sides.

$$\frac{1616}{25}-\frac{128}{5}t+\frac{64}{25}t^{2}-t^{2}=0$$

Combine \frac{64}{25}t^{2} and -t^{2} to get \frac{39}{25}t^{2}.

$$\frac{1616}{25}-\frac{128}{5}t+\frac{39}{25}t^{2}=0$$

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.

$$\frac{39}{25}t^{2}-\frac{128}{5}t+\frac{1616}{25}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{39}{25} for a, -\frac{128}{5} for b, and \frac{1616}{25} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-\left(-\frac{128}{5}\right)±\sqrt{\left(-\frac{128}{5}\right)^{2}-4\times \frac{39}{25}\times \frac{1616}{25}}}{2\times \frac{39}{25}}$$

Square -\frac{128}{5} by squaring both the numerator and the denominator of the fraction.

$$t=\frac{-\left(-\frac{128}{5}\right)±\sqrt{\frac{16384}{25}-4\times \frac{39}{25}\times \frac{1616}{25}}}{2\times \frac{39}{25}}$$

Multiply -4 times \frac{39}{25}.

$$t=\frac{-\left(-\frac{128}{5}\right)±\sqrt{\frac{16384}{25}-\frac{156}{25}\times \frac{1616}{25}}}{2\times \frac{39}{25}}$$

Multiply -\frac{156}{25} times \frac{1616}{25} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$t=\frac{-\left(-\frac{128}{5}\right)±\sqrt{\frac{16384}{25}-\frac{252096}{625}}}{2\times \frac{39}{25}}$$

Add \frac{16384}{25} to -\frac{252096}{625} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$t=\frac{-\left(-\frac{128}{5}\right)±\sqrt{\frac{157504}{625}}}{2\times \frac{39}{25}}$$

Take the square root of \frac{157504}{625}.

$$t=\frac{-\left(-\frac{128}{5}\right)±\frac{8\sqrt{2461}}{25}}{2\times \frac{39}{25}}$$

The opposite of -\frac{128}{5} is \frac{128}{5}.

$$t=\frac{\frac{128}{5}±\frac{8\sqrt{2461}}{25}}{2\times \frac{39}{25}}$$

Multiply 2 times \frac{39}{25}.

$$t=\frac{\frac{128}{5}±\frac{8\sqrt{2461}}{25}}{\frac{78}{25}}$$

Now solve the equation t=\frac{\frac{128}{5}±\frac{8\sqrt{2461}}{25}}{\frac{78}{25}} when ± is plus. Add \frac{128}{5} to \frac{8\sqrt{2461}}{25}.

$$t=\frac{\frac{8\sqrt{2461}}{25}+\frac{128}{5}}{\frac{78}{25}}$$

Divide \frac{128}{5}+\frac{8\sqrt{2461}}{25} by \frac{78}{25} by multiplying \frac{128}{5}+\frac{8\sqrt{2461}}{25} by the reciprocal of \frac{78}{25}.

$$t=\frac{4\sqrt{2461}+320}{39}$$

Now solve the equation t=\frac{\frac{128}{5}±\frac{8\sqrt{2461}}{25}}{\frac{78}{25}} when ± is minus. Subtract \frac{8\sqrt{2461}}{25} from \frac{128}{5}.

$$t=\frac{-\frac{8\sqrt{2461}}{25}+\frac{128}{5}}{\frac{78}{25}}$$

Divide \frac{128}{5}-\frac{8\sqrt{2461}}{25} by \frac{78}{25} by multiplying \frac{128}{5}-\frac{8\sqrt{2461}}{25} by the reciprocal of \frac{78}{25}.

$$t=\frac{320-4\sqrt{2461}}{39}$$

The equation is now solved.

$$t=\frac{4\sqrt{2461}+320}{39} t=\frac{320-4\sqrt{2461}}{39}$$

Substitute \frac{4\sqrt{2461}+320}{39} for t in the equation \sqrt{\left(8-t-\frac{3}{5}t\right)^{2}+\left(\frac{4}{5}\right)^{2}}=t.

$$\sqrt{\left(8-\frac{4\sqrt{2461}+320}{39}-\frac{3}{5}\times \frac{4\sqrt{2461}+320}{39}\right)^{2}+\left(\frac{4}{5}\right)^{2}}=\frac{4\sqrt{2461}+320}{39}$$

Simplify. The value t=\frac{4\sqrt{2461}+320}{39} satisfies the equation.

$$\frac{320}{39}+\frac{4}{39}\times 2461^{\frac{1}{2}}=\frac{4}{39}\times 2461^{\frac{1}{2}}+\frac{320}{39}$$

Substitute \frac{320-4\sqrt{2461}}{39} for t in the equation \sqrt{\left(8-t-\frac{3}{5}t\right)^{2}+\left(\frac{4}{5}\right)^{2}}=t.

$$\sqrt{\left(8-\frac{320-4\sqrt{2461}}{39}-\frac{3}{5}\times \frac{320-4\sqrt{2461}}{39}\right)^{2}+\left(\frac{4}{5}\right)^{2}}=\frac{320-4\sqrt{2461}}{39}$$

Simplify. The value t=\frac{320-4\sqrt{2461}}{39} satisfies the equation.

$$\frac{320}{39}-\frac{4}{39}\times 2461^{\frac{1}{2}}=\frac{320}{39}-\frac{4}{39}\times 2461^{\frac{1}{2}}$$

List all solutions of \sqrt{\left(-\frac{3t}{5}-t+8\right)^{2}+\frac{16}{25}}=t.

$$t=\frac{4\sqrt{2461}+320}{39} t=\frac{320-4\sqrt{2461}}{39}$$