Solve sqrt{3x+20}-4=x

$$\sqrt { 3 x + 20 } - 4 = x$$

Solve for x

$$x=\frac{\sqrt{41}-5}{2}\approx 0.701562119$$

Solution Steps

Subtract -4 from both sides of the equation.

$$\sqrt{3x+20}=x+4$$

Square both sides of the equation.

$$\left(\sqrt{3x+20}\right)^{2}=\left(x+4\right)^{2}$$

Calculate \sqrt{3x+20} to the power of 2 and get 3x+20.

$$3x+20=\left(x+4\right)^{2}$$

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

$$3x+20=x^{2}+8x+16$$

Subtract x^{2} from both sides.

$$3x+20-x^{2}=8x+16$$

Subtract 8x from both sides.

$$3x+20-x^{2}-8x=16$$

Combine 3x and -8x to get -5x.

$$-5x+20-x^{2}=16$$

Subtract 16 from both sides.

$$-5x+20-x^{2}-16=0$$

Subtract 16 from 20 to get 4.

$$-5x+4-x^{2}=0$$

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.

$$-x^{2}-5x+4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -5 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\left(-1\right)\times 4}}{2\left(-1\right)}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\left(-1\right)\times 4}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-5\right)±\sqrt{25+4\times 4}}{2\left(-1\right)}$$

Multiply 4 times 4.

$$x=\frac{-\left(-5\right)±\sqrt{25+16}}{2\left(-1\right)}$$

Add 25 to 16.

$$x=\frac{-\left(-5\right)±\sqrt{41}}{2\left(-1\right)}$$

The opposite of -5 is 5.

$$x=\frac{5±\sqrt{41}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{5±\sqrt{41}}{-2}$$

Now solve the equation x=\frac{5±\sqrt{41}}{-2} when ± is plus. Add 5 to \sqrt{41}.

$$x=\frac{\sqrt{41}+5}{-2}$$

Divide 5+\sqrt{41} by -2.

$$x=\frac{-\sqrt{41}-5}{2}$$

Now solve the equation x=\frac{5±\sqrt{41}}{-2} when ± is minus. Subtract \sqrt{41} from 5.

$$x=\frac{5-\sqrt{41}}{-2}$$

Divide 5-\sqrt{41} by -2.

$$x=\frac{\sqrt{41}-5}{2}$$

The equation is now solved.

$$x=\frac{-\sqrt{41}-5}{2} x=\frac{\sqrt{41}-5}{2}$$

Substitute \frac{-\sqrt{41}-5}{2} for x in the equation \sqrt{3x+20}-4=x.

$$\sqrt{3\times \frac{-\sqrt{41}-5}{2}+20}-4=\frac{-\sqrt{41}-5}{2}$$

Simplify. The value x=\frac{-\sqrt{41}-5}{2} does not satisfy the equation.

$$-\frac{11}{2}+\frac{1}{2}\times 41^{\frac{1}{2}}=-\frac{1}{2}\times 41^{\frac{1}{2}}-\frac{5}{2}$$

Substitute \frac{\sqrt{41}-5}{2} for x in the equation \sqrt{3x+20}-4=x.

$$\sqrt{3\times \frac{\sqrt{41}-5}{2}+20}-4=\frac{\sqrt{41}-5}{2}$$

Simplify. The value x=\frac{\sqrt{41}-5}{2} satisfies the equation.

$$-\frac{5}{2}+\frac{1}{2}\times 41^{\frac{1}{2}}=\frac{1}{2}\times 41^{\frac{1}{2}}-\frac{5}{2}$$

Equation \sqrt{3x+20}=x+4 has a unique solution.

$$x=\frac{\sqrt{41}-5}{2}$$