Solve sqrt{1-2x/3}=x+5

$$\sqrt { 1 - \frac { 2 x } { 3 } } = x + 5$$

Solve for x

$$x=\frac{2\left(\sqrt{10}-8\right)}{3}\approx -3.225148227$$

Solution Steps

Square both sides of the equation.

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

Calculate \sqrt{1-\frac{2x}{3}} to the power of 2 and get 1-\frac{2x}{3}.

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

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

$$1-\frac{2x}{3}=x^{2}+10x+25$$

Multiply both sides of the equation by 3.

$$3-2x=3x^{2}+30x+75$$

Subtract 3x^{2} from both sides.

$$3-2x-3x^{2}=30x+75$$

Subtract 30x from both sides.

$$3-2x-3x^{2}-30x=75$$

Combine -2x and -30x to get -32x.

$$3-32x-3x^{2}=75$$

Subtract 75 from both sides.

$$3-32x-3x^{2}-75=0$$

Subtract 75 from 3 to get -72.

$$-72-32x-3x^{2}=0$$

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.

$$-3x^{2}-32x-72=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, -32 for b, and -72 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-32\right)±\sqrt{\left(-32\right)^{2}-4\left(-3\right)\left(-72\right)}}{2\left(-3\right)}$$

Square -32.

$$x=\frac{-\left(-32\right)±\sqrt{1024-4\left(-3\right)\left(-72\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-\left(-32\right)±\sqrt{1024+12\left(-72\right)}}{2\left(-3\right)}$$

Multiply 12 times -72.

$$x=\frac{-\left(-32\right)±\sqrt{1024-864}}{2\left(-3\right)}$$

Add 1024 to -864.

$$x=\frac{-\left(-32\right)±\sqrt{160}}{2\left(-3\right)}$$

Take the square root of 160.

$$x=\frac{-\left(-32\right)±4\sqrt{10}}{2\left(-3\right)}$$

The opposite of -32 is 32.

$$x=\frac{32±4\sqrt{10}}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{32±4\sqrt{10}}{-6}$$

Now solve the equation x=\frac{32±4\sqrt{10}}{-6} when ± is plus. Add 32 to 4\sqrt{10}.

$$x=\frac{4\sqrt{10}+32}{-6}$$

Divide 32+4\sqrt{10} by -6.

$$x=\frac{-2\sqrt{10}-16}{3}$$

Now solve the equation x=\frac{32±4\sqrt{10}}{-6} when ± is minus. Subtract 4\sqrt{10} from 32.

$$x=\frac{32-4\sqrt{10}}{-6}$$

Divide 32-4\sqrt{10} by -6.

$$x=\frac{2\sqrt{10}-16}{3}$$

The equation is now solved.

$$x=\frac{-2\sqrt{10}-16}{3} x=\frac{2\sqrt{10}-16}{3}$$

Substitute \frac{-2\sqrt{10}-16}{3} for x in the equation \sqrt{1-\frac{2x}{3}}=x+5.

$$\sqrt{1-\frac{2\times \frac{-2\sqrt{10}-16}{3}}{3}}=\frac{-2\sqrt{10}-16}{3}+5$$

Simplify. The value x=\frac{-2\sqrt{10}-16}{3} does not satisfy the equation because the left and the right hand side have opposite signs.

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

Substitute \frac{2\sqrt{10}-16}{3} for x in the equation \sqrt{1-\frac{2x}{3}}=x+5.

$$\sqrt{1-\frac{2\times \frac{2\sqrt{10}-16}{3}}{3}}=\frac{2\sqrt{10}-16}{3}+5$$

Simplify. The value x=\frac{2\sqrt{10}-16}{3} satisfies the equation.

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

Equation \sqrt{-\frac{2x}{3}+1}=x+5 has a unique solution.

$$x=\frac{2\sqrt{10}-16}{3}$$