Solve sqrt{x-2/3}=25-4x/5

$$\sqrt{ \frac{ x-2 }{ 3 } } = \frac{ 25-4x }{ 5 }$$

Solve for x

$$x=5$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{\frac{x-2}{3}}\right)^{2}=\left(\frac{25-4x}{5}\right)^{2}$$

Calculate \sqrt{\frac{x-2}{3}} to the power of 2 and get \frac{x-2}{3}.

$$\frac{x-2}{3}=\left(\frac{25-4x}{5}\right)^{2}$$

To raise \frac{25-4x}{5} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{x-2}{3}=\frac{\left(25-4x\right)^{2}}{5^{2}}$$

Divide each term of x-2 by 3 to get \frac{1}{3}x-\frac{2}{3}.

$$\frac{1}{3}x-\frac{2}{3}=\frac{\left(25-4x\right)^{2}}{5^{2}}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(25-4x\right)^{2}.

$$\frac{1}{3}x-\frac{2}{3}=\frac{625-200x+16x^{2}}{5^{2}}$$

Calculate 5 to the power of 2 and get 25.

$$\frac{1}{3}x-\frac{2}{3}=\frac{625-200x+16x^{2}}{25}$$

Divide each term of 625-200x+16x^{2} by 25 to get 25-8x+\frac{16}{25}x^{2}.

$$\frac{1}{3}x-\frac{2}{3}=25-8x+\frac{16}{25}x^{2}$$

Subtract 25 from both sides.

$$\frac{1}{3}x-\frac{2}{3}-25=-8x+\frac{16}{25}x^{2}$$

Subtract 25 from -\frac{2}{3} to get -\frac{77}{3}.

$$\frac{1}{3}x-\frac{77}{3}=-8x+\frac{16}{25}x^{2}$$

Add 8x to both sides.

$$\frac{1}{3}x-\frac{77}{3}+8x=\frac{16}{25}x^{2}$$

Combine \frac{1}{3}x and 8x to get \frac{25}{3}x.

$$\frac{25}{3}x-\frac{77}{3}=\frac{16}{25}x^{2}$$

Subtract \frac{16}{25}x^{2} from both sides.

$$\frac{25}{3}x-\frac{77}{3}-\frac{16}{25}x^{2}=0$$

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.

$$-\frac{16}{25}x^{2}+\frac{25}{3}x-\frac{77}{3}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{16}{25} for a, \frac{25}{3} for b, and -\frac{77}{3} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\frac{25}{3}±\sqrt{\left(\frac{25}{3}\right)^{2}-4\left(-\frac{16}{25}\right)\left(-\frac{77}{3}\right)}}{2\left(-\frac{16}{25}\right)}$$

Square \frac{25}{3} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\frac{25}{3}±\sqrt{\frac{625}{9}-4\left(-\frac{16}{25}\right)\left(-\frac{77}{3}\right)}}{2\left(-\frac{16}{25}\right)}$$

Multiply -4 times -\frac{16}{25}.

$$x=\frac{-\frac{25}{3}±\sqrt{\frac{625}{9}+\frac{64}{25}\left(-\frac{77}{3}\right)}}{2\left(-\frac{16}{25}\right)}$$

Multiply \frac{64}{25} times -\frac{77}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\frac{25}{3}±\sqrt{\frac{625}{9}-\frac{4928}{75}}}{2\left(-\frac{16}{25}\right)}$$

Add \frac{625}{9} to -\frac{4928}{75} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\frac{25}{3}±\sqrt{\frac{841}{225}}}{2\left(-\frac{16}{25}\right)}$$

Take the square root of \frac{841}{225}.

$$x=\frac{-\frac{25}{3}±\frac{29}{15}}{2\left(-\frac{16}{25}\right)}$$

Multiply 2 times -\frac{16}{25}.

$$x=\frac{-\frac{25}{3}±\frac{29}{15}}{-\frac{32}{25}}$$

Now solve the equation x=\frac{-\frac{25}{3}±\frac{29}{15}}{-\frac{32}{25}} when ± is plus. Add -\frac{25}{3} to \frac{29}{15} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{\frac{32}{5}}{-\frac{32}{25}}$$

Divide -\frac{32}{5} by -\frac{32}{25} by multiplying -\frac{32}{5} by the reciprocal of -\frac{32}{25}.

$$x=5$$

Now solve the equation x=\frac{-\frac{25}{3}±\frac{29}{15}}{-\frac{32}{25}} when ± is minus. Subtract \frac{29}{15} from -\frac{25}{3} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{\frac{154}{15}}{-\frac{32}{25}}$$

Divide -\frac{154}{15} by -\frac{32}{25} by multiplying -\frac{154}{15} by the reciprocal of -\frac{32}{25}.

$$x=\frac{385}{48}$$

The equation is now solved.

$$x=5 x=\frac{385}{48}$$

Substitute 5 for x in the equation \sqrt{\frac{x-2}{3}}=\frac{25-4x}{5}.

$$\sqrt{\frac{5-2}{3}}=\frac{25-4\times 5}{5}$$

Simplify. The value x=5 satisfies the equation.

$$1=1$$

Substitute \frac{385}{48} for x in the equation \sqrt{\frac{x-2}{3}}=\frac{25-4x}{5}.

$$\sqrt{\frac{\frac{385}{48}-2}{3}}=\frac{25-4\times \frac{385}{48}}{5}$$

Simplify. The value x=\frac{385}{48} does not satisfy the equation because the left and the right hand side have opposite signs.

$$\frac{17}{12}=-\frac{17}{12}$$

Equation \sqrt{\frac{x-2}{3}}=\frac{25-4x}{5} has a unique solution.

$$x=5$$