Solve sqrt{5x+7}-2=x

$$\sqrt { 5 x + 7 } - 2 = x$$

Solve for x

$$x = \frac{\sqrt{13} + 1}{2} \approx 2.302775638$$

$$x=\frac{1-\sqrt{13}}{2}\approx -1.302775638$$

Solution Steps

Subtract -2 from both sides of the equation.

$$\sqrt{5x+7}=x+2$$

Square both sides of the equation.

$$\left(\sqrt{5x+7}\right)^{2}=\left(x+2\right)^{2}$$

Calculate \sqrt{5x+7} to the power of 2 and get 5x+7.

$$5x+7=\left(x+2\right)^{2}$$

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

$$5x+7=x^{2}+4x+4$$

Subtract x^{2} from both sides.

$$5x+7-x^{2}=4x+4$$

Subtract 4x from both sides.

$$5x+7-x^{2}-4x=4$$

Combine 5x and -4x to get x.

$$x+7-x^{2}=4$$

Subtract 4 from both sides.

$$x+7-x^{2}-4=0$$

Subtract 4 from 7 to get 3.

$$x+3-x^{2}=0$$

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.

$$-x^{2}+x+3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 1 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1±\sqrt{1^{2}-4\left(-1\right)\times 3}}{2\left(-1\right)}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\left(-1\right)\times 3}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-1±\sqrt{1+4\times 3}}{2\left(-1\right)}$$

Multiply 4 times 3.

$$x=\frac{-1±\sqrt{1+12}}{2\left(-1\right)}$$

Add 1 to 12.

$$x=\frac{-1±\sqrt{13}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-1±\sqrt{13}}{-2}$$

Now solve the equation x=\frac{-1±\sqrt{13}}{-2} when ± is plus. Add -1 to \sqrt{13}.

$$x=\frac{\sqrt{13}-1}{-2}$$

Divide -1+\sqrt{13} by -2.

$$x=\frac{1-\sqrt{13}}{2}$$

Now solve the equation x=\frac{-1±\sqrt{13}}{-2} when ± is minus. Subtract \sqrt{13} from -1.

$$x=\frac{-\sqrt{13}-1}{-2}$$

Divide -1-\sqrt{13} by -2.

$$x=\frac{\sqrt{13}+1}{2}$$

The equation is now solved.

$$x=\frac{1-\sqrt{13}}{2} x=\frac{\sqrt{13}+1}{2}$$

Substitute \frac{1-\sqrt{13}}{2} for x in the equation \sqrt{5x+7}-2=x.

$$\sqrt{5\times \frac{1-\sqrt{13}}{2}+7}-2=\frac{1-\sqrt{13}}{2}$$

Simplify. The value x=\frac{1-\sqrt{13}}{2} satisfies the equation.

$$\frac{1}{2}-\frac{1}{2}\times 13^{\frac{1}{2}}=\frac{1}{2}-\frac{1}{2}\times 13^{\frac{1}{2}}$$

Substitute \frac{\sqrt{13}+1}{2} for x in the equation \sqrt{5x+7}-2=x.

$$\sqrt{5\times \frac{\sqrt{13}+1}{2}+7}-2=\frac{\sqrt{13}+1}{2}$$

Simplify. The value x=\frac{\sqrt{13}+1}{2} satisfies the equation.

$$\frac{1}{2}+\frac{1}{2}\times 13^{\frac{1}{2}}=\frac{1}{2}\times 13^{\frac{1}{2}}+\frac{1}{2}$$

List all solutions of \sqrt{5x+7}=x+2.

$$x=\frac{1-\sqrt{13}}{2} x=\frac{\sqrt{13}+1}{2}$$