Solve sqrt{3x-7}=x-5

$$\sqrt { 3 x - 7 } = x - 5$$

Solve for x

$$x = \frac{\sqrt{41} + 13}{2} \approx 9.701562119$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{3x-7}\right)^{2}=\left(x-5\right)^{2}$$

Calculate \sqrt{3x-7} to the power of 2 and get 3x-7.

$$3x-7=\left(x-5\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-5\right)^{2}.

$$3x-7=x^{2}-10x+25$$

Subtract x^{2} from both sides.

$$3x-7-x^{2}=-10x+25$$

Add 10x to both sides.

$$3x-7-x^{2}+10x=25$$

Combine 3x and 10x to get 13x.

$$13x-7-x^{2}=25$$

Subtract 25 from both sides.

$$13x-7-x^{2}-25=0$$

Subtract 25 from -7 to get -32.

$$13x-32-x^{2}=0$$

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.

$$-x^{2}+13x-32=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 13 for b, and -32 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-13±\sqrt{13^{2}-4\left(-1\right)\left(-32\right)}}{2\left(-1\right)}$$

Square 13.

$$x=\frac{-13±\sqrt{169-4\left(-1\right)\left(-32\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-13±\sqrt{169+4\left(-32\right)}}{2\left(-1\right)}$$

Multiply 4 times -32.

$$x=\frac{-13±\sqrt{169-128}}{2\left(-1\right)}$$

Add 169 to -128.

$$x=\frac{-13±\sqrt{41}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-13±\sqrt{41}}{-2}$$

Now solve the equation x=\frac{-13±\sqrt{41}}{-2} when ± is plus. Add -13 to \sqrt{41}.

$$x=\frac{\sqrt{41}-13}{-2}$$

Divide -13+\sqrt{41} by -2.

$$x=\frac{13-\sqrt{41}}{2}$$

Now solve the equation x=\frac{-13±\sqrt{41}}{-2} when ± is minus. Subtract \sqrt{41} from -13.

$$x=\frac{-\sqrt{41}-13}{-2}$$

Divide -13-\sqrt{41} by -2.

$$x=\frac{\sqrt{41}+13}{2}$$

The equation is now solved.

$$x=\frac{13-\sqrt{41}}{2} x=\frac{\sqrt{41}+13}{2}$$

Substitute \frac{13-\sqrt{41}}{2} for x in the equation \sqrt{3x-7}=x-5.

$$\sqrt{3\times \frac{13-\sqrt{41}}{2}-7}=\frac{13-\sqrt{41}}{2}-5$$

Simplify. The value x=\frac{13-\sqrt{41}}{2} does not satisfy the equation because the left and the right hand side have opposite signs.

$$-\left(\frac{3}{2}-\frac{1}{2}\times 41^{\frac{1}{2}}\right)=\frac{3}{2}-\frac{1}{2}\times 41^{\frac{1}{2}}$$

Substitute \frac{\sqrt{41}+13}{2} for x in the equation \sqrt{3x-7}=x-5.

$$\sqrt{3\times \frac{\sqrt{41}+13}{2}-7}=\frac{\sqrt{41}+13}{2}-5$$

Simplify. The value x=\frac{\sqrt{41}+13}{2} satisfies the equation.

$$\frac{3}{2}+\frac{1}{2}\times 41^{\frac{1}{2}}=\frac{1}{2}\times 41^{\frac{1}{2}}+\frac{3}{2}$$

Equation \sqrt{3x-7}=x-5 has a unique solution.

$$x=\frac{\sqrt{41}+13}{2}$$