Solve sqrt{x-3}=x+3

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

Solve for x (complex solution)

$$x=\frac{-5+\sqrt{23}i}{2}\approx -2.5+2.397915762i$$

Solution Steps

Square both sides of the equation.

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

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

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

$$x-3=x^{2}+6x+9$$

Subtract x^{2} from both sides.

$$x-3-x^{2}=6x+9$$

Subtract 6x from both sides.

$$x-3-x^{2}-6x=9$$

Combine x and -6x to get -5x.

$$-5x-3-x^{2}=9$$

Subtract 9 from both sides.

$$-5x-3-x^{2}-9=0$$

Subtract 9 from -3 to get -12.

$$-5x-12-x^{2}=0$$

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.

$$-x^{2}-5x-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -5 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\left(-1\right)\left(-12\right)}}{2\left(-1\right)}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\left(-1\right)\left(-12\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-5\right)±\sqrt{25+4\left(-12\right)}}{2\left(-1\right)}$$

Multiply 4 times -12.

$$x=\frac{-\left(-5\right)±\sqrt{25-48}}{2\left(-1\right)}$$

Add 25 to -48.

$$x=\frac{-\left(-5\right)±\sqrt{-23}}{2\left(-1\right)}$$

Take the square root of -23.

$$x=\frac{-\left(-5\right)±\sqrt{23}i}{2\left(-1\right)}$$

The opposite of -5 is 5.

$$x=\frac{5±\sqrt{23}i}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{5±\sqrt{23}i}{-2}$$

Now solve the equation x=\frac{5±\sqrt{23}i}{-2} when ± is plus. Add 5 to i\sqrt{23}.

$$x=\frac{5+\sqrt{23}i}{-2}$$

Divide 5+i\sqrt{23} by -2.

$$x=\frac{-\sqrt{23}i-5}{2}$$

Now solve the equation x=\frac{5±\sqrt{23}i}{-2} when ± is minus. Subtract i\sqrt{23} from 5.

$$x=\frac{-\sqrt{23}i+5}{-2}$$

Divide 5-i\sqrt{23} by -2.

$$x=\frac{-5+\sqrt{23}i}{2}$$

The equation is now solved.

$$x=\frac{-\sqrt{23}i-5}{2} x=\frac{-5+\sqrt{23}i}{2}$$

Substitute \frac{-\sqrt{23}i-5}{2} for x in the equation \sqrt{x-3}=x+3.

$$\sqrt{\frac{-\sqrt{23}i-5}{2}-3}=\frac{-\sqrt{23}i-5}{2}+3$$

Simplify. The value x=\frac{-\sqrt{23}i-5}{2} does not satisfy the equation.

$$-\left(\frac{1}{2}-\frac{1}{2}i\times 23^{\frac{1}{2}}\right)=-\frac{1}{2}i\times 23^{\frac{1}{2}}+\frac{1}{2}$$

Substitute \frac{-5+\sqrt{23}i}{2} for x in the equation \sqrt{x-3}=x+3.

$$\sqrt{\frac{-5+\sqrt{23}i}{2}-3}=\frac{-5+\sqrt{23}i}{2}+3$$

Simplify. The value x=\frac{-5+\sqrt{23}i}{2} satisfies the equation.

$$\frac{1}{2}+\frac{1}{2}i\times 23^{\frac{1}{2}}=\frac{1}{2}+\frac{1}{2}i\times 23^{\frac{1}{2}}$$

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

$$x=\frac{-5+\sqrt{23}i}{2}$$