Solve sqrt{x+5}=x+4

$$\sqrt{ x+5 } = x+4$$

Solve for x

$$x=\frac{\sqrt{5}-7}{2}\approx -2.381966011$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{x+5}\right)^{2}=\left(x+4\right)^{2}$$

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

$$x+5=\left(x+4\right)^{2}$$

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

$$x+5=x^{2}+8x+16$$

Subtract x^{2} from both sides.

$$x+5-x^{2}=8x+16$$

Subtract 8x from both sides.

$$x+5-x^{2}-8x=16$$

Combine x and -8x to get -7x.

$$-7x+5-x^{2}=16$$

Subtract 16 from both sides.

$$-7x+5-x^{2}-16=0$$

Subtract 16 from 5 to get -11.

$$-7x-11-x^{2}=0$$

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.

$$-x^{2}-7x-11=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -7 for b, and -11 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\left(-1\right)\left(-11\right)}}{2\left(-1\right)}$$

Square -7.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\left(-1\right)\left(-11\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-7\right)±\sqrt{49+4\left(-11\right)}}{2\left(-1\right)}$$

Multiply 4 times -11.

$$x=\frac{-\left(-7\right)±\sqrt{49-44}}{2\left(-1\right)}$$

Add 49 to -44.

$$x=\frac{-\left(-7\right)±\sqrt{5}}{2\left(-1\right)}$$

The opposite of -7 is 7.

$$x=\frac{7±\sqrt{5}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{7±\sqrt{5}}{-2}$$

Now solve the equation x=\frac{7±\sqrt{5}}{-2} when ± is plus. Add 7 to \sqrt{5}.

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

Divide 7+\sqrt{5} by -2.

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

Now solve the equation x=\frac{7±\sqrt{5}}{-2} when ± is minus. Subtract \sqrt{5} from 7.

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

Divide 7-\sqrt{5} by -2.

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

The equation is now solved.

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

Substitute \frac{-\sqrt{5}-7}{2} for x in the equation \sqrt{x+5}=x+4.

$$\sqrt{\frac{-\sqrt{5}-7}{2}+5}=\frac{-\sqrt{5}-7}{2}+4$$

Simplify. The value x=\frac{-\sqrt{5}-7}{2} does not satisfy the equation because the left and the right hand side have opposite signs.

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

Substitute \frac{\sqrt{5}-7}{2} for x in the equation \sqrt{x+5}=x+4.

$$\sqrt{\frac{\sqrt{5}-7}{2}+5}=\frac{\sqrt{5}-7}{2}+4$$

Simplify. The value x=\frac{\sqrt{5}-7}{2} satisfies the equation.

$$\frac{1}{2}+\frac{1}{2}\times 5^{\frac{1}{2}}=\frac{1}{2}\times 5^{\frac{1}{2}}+\frac{1}{2}$$

Equation \sqrt{x+5}=x+4 has a unique solution.

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