Solve sqrt{x-2}=11-x/2

$$\sqrt { x - 2 } = \frac { 11 - x } { 2 }$$

Solve for x

$$x=13-2\sqrt{10}\approx 6.67544468$$

Solution Steps

Square both sides of the equation.

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

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

$$x-2=\left(\frac{11-x}{2}\right)^{2}$$

To raise \frac{11-x}{2} to a power, raise both numerator and denominator to the power and then divide.

$$x-2=\frac{\left(11-x\right)^{2}}{2^{2}}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(11-x\right)^{2}.

$$x-2=\frac{121-22x+x^{2}}{2^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$x-2=\frac{121-22x+x^{2}}{4}$$

Divide each term of 121-22x+x^{2} by 4 to get \frac{121}{4}-\frac{11}{2}x+\frac{1}{4}x^{2}.

$$x-2=\frac{121}{4}-\frac{11}{2}x+\frac{1}{4}x^{2}$$

Subtract \frac{121}{4} from both sides.

$$x-2-\frac{121}{4}=-\frac{11}{2}x+\frac{1}{4}x^{2}$$

Subtract \frac{121}{4} from -2 to get -\frac{129}{4}.

$$x-\frac{129}{4}=-\frac{11}{2}x+\frac{1}{4}x^{2}$$

Add \frac{11}{2}x to both sides.

$$x-\frac{129}{4}+\frac{11}{2}x=\frac{1}{4}x^{2}$$

Combine x and \frac{11}{2}x to get \frac{13}{2}x.

$$\frac{13}{2}x-\frac{129}{4}=\frac{1}{4}x^{2}$$

Subtract \frac{1}{4}x^{2} from both sides.

$$\frac{13}{2}x-\frac{129}{4}-\frac{1}{4}x^{2}=0$$

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.

$$-\frac{1}{4}x^{2}+\frac{13}{2}x-\frac{129}{4}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{1}{4} for a, \frac{13}{2} for b, and -\frac{129}{4} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\frac{13}{2}±\sqrt{\left(\frac{13}{2}\right)^{2}-4\left(-\frac{1}{4}\right)\left(-\frac{129}{4}\right)}}{2\left(-\frac{1}{4}\right)}$$

Square \frac{13}{2} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\frac{13}{2}±\sqrt{\frac{169}{4}-4\left(-\frac{1}{4}\right)\left(-\frac{129}{4}\right)}}{2\left(-\frac{1}{4}\right)}$$

Multiply -4 times -\frac{1}{4}.

$$x=\frac{-\frac{13}{2}±\sqrt{\frac{169-129}{4}}}{2\left(-\frac{1}{4}\right)}$$

Add \frac{169}{4} to -\frac{129}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\frac{13}{2}±\sqrt{10}}{2\left(-\frac{1}{4}\right)}$$

Multiply 2 times -\frac{1}{4}.

$$x=\frac{-\frac{13}{2}±\sqrt{10}}{-\frac{1}{2}}$$

Now solve the equation x=\frac{-\frac{13}{2}±\sqrt{10}}{-\frac{1}{2}} when ± is plus. Add -\frac{13}{2} to \sqrt{10}.

$$x=\frac{\sqrt{10}-\frac{13}{2}}{-\frac{1}{2}}$$

Divide -\frac{13}{2}+\sqrt{10} by -\frac{1}{2} by multiplying -\frac{13}{2}+\sqrt{10} by the reciprocal of -\frac{1}{2}.

$$x=13-2\sqrt{10}$$

Now solve the equation x=\frac{-\frac{13}{2}±\sqrt{10}}{-\frac{1}{2}} when ± is minus. Subtract \sqrt{10} from -\frac{13}{2}.

$$x=\frac{-\sqrt{10}-\frac{13}{2}}{-\frac{1}{2}}$$

Divide -\frac{13}{2}-\sqrt{10} by -\frac{1}{2} by multiplying -\frac{13}{2}-\sqrt{10} by the reciprocal of -\frac{1}{2}.

$$x=2\sqrt{10}+13$$

The equation is now solved.

$$x=13-2\sqrt{10} x=2\sqrt{10}+13$$

Substitute 13-2\sqrt{10} for x in the equation \sqrt{x-2}=\frac{11-x}{2}.

$$\sqrt{13-2\sqrt{10}-2}=\frac{11-\left(13-2\sqrt{10}\right)}{2}$$

Simplify. The value x=13-2\sqrt{10} satisfies the equation.

$$10^{\frac{1}{2}}-1=-1+10^{\frac{1}{2}}$$

Substitute 2\sqrt{10}+13 for x in the equation \sqrt{x-2}=\frac{11-x}{2}.

$$\sqrt{2\sqrt{10}+13-2}=\frac{11-\left(2\sqrt{10}+13\right)}{2}$$

Simplify. The value x=2\sqrt{10}+13 does not satisfy the equation because the left and the right hand side have opposite signs.

$$10^{\frac{1}{2}}+1=-1-10^{\frac{1}{2}}$$

Equation \sqrt{x-2}=\frac{11-x}{2} has a unique solution.

$$x=13-2\sqrt{10}$$