Solve sqrt{2x+7}-3x+1=x-1

$$\sqrt { 2 x + 7 } - 3 x + 1 = x - 1$$

Solve for x

$$x = \frac{\sqrt{129} + 9}{16} \approx 1.272363543$$

Solution Steps

Subtract -3x+1 from both sides of the equation.

$$\sqrt{2x+7}=x-1-\left(-3x+1\right)$$

To find the opposite of -3x+1, find the opposite of each term.

$$\sqrt{2x+7}=x-1-\left(-3x\right)-1$$

The opposite of -3x is 3x.

$$\sqrt{2x+7}=x-1+3x-1$$

Combine x and 3x to get 4x.

$$\sqrt{2x+7}=4x-1-1$$

Subtract 1 from -1 to get -2.

$$\sqrt{2x+7}=4x-2$$

Square both sides of the equation.

$$\left(\sqrt{2x+7}\right)^{2}=\left(4x-2\right)^{2}$$

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

$$2x+7=\left(4x-2\right)^{2}$$

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

$$2x+7=16x^{2}-16x+4$$

Subtract 16x^{2} from both sides.

$$2x+7-16x^{2}=-16x+4$$

Add 16x to both sides.

$$2x+7-16x^{2}+16x=4$$

Combine 2x and 16x to get 18x.

$$18x+7-16x^{2}=4$$

Subtract 4 from both sides.

$$18x+7-16x^{2}-4=0$$

Subtract 4 from 7 to get 3.

$$18x+3-16x^{2}=0$$

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.

$$-16x^{2}+18x+3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -16 for a, 18 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-18±\sqrt{18^{2}-4\left(-16\right)\times 3}}{2\left(-16\right)}$$

Square 18.

$$x=\frac{-18±\sqrt{324-4\left(-16\right)\times 3}}{2\left(-16\right)}$$

Multiply -4 times -16.

$$x=\frac{-18±\sqrt{324+64\times 3}}{2\left(-16\right)}$$

Multiply 64 times 3.

$$x=\frac{-18±\sqrt{324+192}}{2\left(-16\right)}$$

Add 324 to 192.

$$x=\frac{-18±\sqrt{516}}{2\left(-16\right)}$$

Take the square root of 516.

$$x=\frac{-18±2\sqrt{129}}{2\left(-16\right)}$$

Multiply 2 times -16.

$$x=\frac{-18±2\sqrt{129}}{-32}$$

Now solve the equation x=\frac{-18±2\sqrt{129}}{-32} when ± is plus. Add -18 to 2\sqrt{129}.

$$x=\frac{2\sqrt{129}-18}{-32}$$

Divide -18+2\sqrt{129} by -32.

$$x=\frac{9-\sqrt{129}}{16}$$

Now solve the equation x=\frac{-18±2\sqrt{129}}{-32} when ± is minus. Subtract 2\sqrt{129} from -18.

$$x=\frac{-2\sqrt{129}-18}{-32}$$

Divide -18-2\sqrt{129} by -32.

$$x=\frac{\sqrt{129}+9}{16}$$

The equation is now solved.

$$x=\frac{9-\sqrt{129}}{16} x=\frac{\sqrt{129}+9}{16}$$

Substitute \frac{9-\sqrt{129}}{16} for x in the equation \sqrt{2x+7}-3x+1=x-1.

$$\sqrt{2\times \frac{9-\sqrt{129}}{16}+7}-3\times \frac{9-\sqrt{129}}{16}+1=\frac{9-\sqrt{129}}{16}-1$$

Simplify. The value x=\frac{9-\sqrt{129}}{16} does not satisfy the equation because the left and the right hand side have opposite signs.

$$-\frac{15}{16}+\frac{7}{16}\times 129^{\frac{1}{2}}=-\frac{7}{16}-\frac{1}{16}\times 129^{\frac{1}{2}}$$

Substitute \frac{\sqrt{129}+9}{16} for x in the equation \sqrt{2x+7}-3x+1=x-1.

$$\sqrt{2\times \frac{\sqrt{129}+9}{16}+7}-3\times \frac{\sqrt{129}+9}{16}+1=\frac{\sqrt{129}+9}{16}-1$$

Simplify. The value x=\frac{\sqrt{129}+9}{16} satisfies the equation.

$$-\frac{7}{16}+\frac{1}{16}\times 129^{\frac{1}{2}}=\frac{1}{16}\times 129^{\frac{1}{2}}-\frac{7}{16}$$

Equation \sqrt{2x+7}=4x-2 has a unique solution.

$$x=\frac{\sqrt{129}+9}{16}$$