Solve sqrt{2x-55}=x+3

$$\sqrt { 2 x - 55 } = x + 3$$

Solve for x (complex solution)

$$x=-2+2\sqrt{15}i\approx -2+7.745966692i$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{2x-55}\right)^{2}=\left(x+3\right)^{2}$$

Calculate \sqrt{2x-55} to the power of 2 and get 2x-55.

$$2x-55=\left(x+3\right)^{2}$$

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

$$2x-55=x^{2}+6x+9$$

Subtract x^{2} from both sides.

$$2x-55-x^{2}=6x+9$$

Subtract 6x from both sides.

$$2x-55-x^{2}-6x=9$$

Combine 2x and -6x to get -4x.

$$-4x-55-x^{2}=9$$

Subtract 9 from both sides.

$$-4x-55-x^{2}-9=0$$

Subtract 9 from -55 to get -64.

$$-4x-64-x^{2}=0$$

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.

$$-x^{2}-4x-64=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -4 for b, and -64 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\left(-1\right)\left(-64\right)}}{2\left(-1\right)}$$

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\left(-1\right)\left(-64\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-4\right)±\sqrt{16+4\left(-64\right)}}{2\left(-1\right)}$$

Multiply 4 times -64.

$$x=\frac{-\left(-4\right)±\sqrt{16-256}}{2\left(-1\right)}$$

Add 16 to -256.

$$x=\frac{-\left(-4\right)±\sqrt{-240}}{2\left(-1\right)}$$

Take the square root of -240.

$$x=\frac{-\left(-4\right)±4\sqrt{15}i}{2\left(-1\right)}$$

The opposite of -4 is 4.

$$x=\frac{4±4\sqrt{15}i}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{4±4\sqrt{15}i}{-2}$$

Now solve the equation x=\frac{4±4\sqrt{15}i}{-2} when ± is plus. Add 4 to 4i\sqrt{15}.

$$x=\frac{4+4\sqrt{15}i}{-2}$$

Divide 4+4i\sqrt{15} by -2.

$$x=-2\sqrt{15}i-2$$

Now solve the equation x=\frac{4±4\sqrt{15}i}{-2} when ± is minus. Subtract 4i\sqrt{15} from 4.

$$x=\frac{-4\sqrt{15}i+4}{-2}$$

Divide 4-4i\sqrt{15} by -2.

$$x=-2+2\sqrt{15}i$$

The equation is now solved.

$$x=-2\sqrt{15}i-2 x=-2+2\sqrt{15}i$$

Substitute -2\sqrt{15}i-2 for x in the equation \sqrt{2x-55}=x+3.

$$\sqrt{2\left(-2\sqrt{15}i-2\right)-55}=-2\sqrt{15}i-2+3$$

Simplify. The value x=-2\sqrt{15}i-2 does not satisfy the equation.

$$-\left(1-2i\times 15^{\frac{1}{2}}\right)=-2i\times 15^{\frac{1}{2}}+1$$

Substitute -2+2\sqrt{15}i for x in the equation \sqrt{2x-55}=x+3.

$$\sqrt{2\left(-2+2\sqrt{15}i\right)-55}=-2+2\sqrt{15}i+3$$

Simplify. The value x=-2+2\sqrt{15}i satisfies the equation.

$$1+2i\times 15^{\frac{1}{2}}=1+2i\times 15^{\frac{1}{2}}$$

Equation \sqrt{2x-55}=x+3 has a unique solution.

$$x=-2+2\sqrt{15}i$$