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

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

Solve for x

$$x=2-\sqrt{6}\approx -0.449489743$$

Solution Steps

Square both sides of the equation.

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

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

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(1-x\right)^{2}.

$$2x+3=1-2x+x^{2}$$

Subtract 1 from both sides.

$$2x+3-1=-2x+x^{2}$$

Subtract 1 from 3 to get 2.

$$2x+2=-2x+x^{2}$$

Add 2x to both sides.

$$2x+2+2x=x^{2}$$

Combine 2x and 2x to get 4x.

$$4x+2=x^{2}$$

Subtract x^{2} from both sides.

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

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.

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

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

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

Square 4.

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

Multiply -4 times -1.

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

Multiply 4 times 2.

$$x=\frac{-4±\sqrt{16+8}}{2\left(-1\right)}$$

Add 16 to 8.

$$x=\frac{-4±\sqrt{24}}{2\left(-1\right)}$$

Take the square root of 24.

$$x=\frac{-4±2\sqrt{6}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-4±2\sqrt{6}}{-2}$$

Now solve the equation x=\frac{-4±2\sqrt{6}}{-2} when ± is plus. Add -4 to 2\sqrt{6}.

$$x=\frac{2\sqrt{6}-4}{-2}$$

Divide -4+2\sqrt{6} by -2.

$$x=2-\sqrt{6}$$

Now solve the equation x=\frac{-4±2\sqrt{6}}{-2} when ± is minus. Subtract 2\sqrt{6} from -4.

$$x=\frac{-2\sqrt{6}-4}{-2}$$

Divide -4-2\sqrt{6} by -2.

$$x=\sqrt{6}+2$$

The equation is now solved.

$$x=2-\sqrt{6} x=\sqrt{6}+2$$

Substitute 2-\sqrt{6} for x in the equation \sqrt{2x+3}=1-x.

$$\sqrt{2\left(2-\sqrt{6}\right)+3}=1-\left(2-\sqrt{6}\right)$$

Simplify. The value x=2-\sqrt{6} satisfies the equation.

$$6^{\frac{1}{2}}-1=-1+6^{\frac{1}{2}}$$

Substitute \sqrt{6}+2 for x in the equation \sqrt{2x+3}=1-x.

$$\sqrt{2\left(\sqrt{6}+2\right)+3}=1-\left(\sqrt{6}+2\right)$$

Simplify. The value x=\sqrt{6}+2 does not satisfy the equation because the left and the right hand side have opposite signs.

$$6^{\frac{1}{2}}+1=-1-6^{\frac{1}{2}}$$

Equation \sqrt{2x+3}=1-x has a unique solution.

$$x=2-\sqrt{6}$$