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

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

Solve for x

$$x=\frac{1}{2}=0.5$$

Solution Steps

Subtract 2 from both sides of the equation.

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

Subtract 2 from -1 to get -3.

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

Square both sides of the equation.

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

Expand \left(-\sqrt{2x+3}\right)^{2}.

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

Calculate -1 to the power of 2 and get 1.

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

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

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

Use the distributive property to multiply 1 by 2x+3.

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

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

$$2x+3=4x^{2}-12x+9$$

Subtract 4x^{2} from both sides.

$$2x+3-4x^{2}=-12x+9$$

Add 12x to both sides.

$$2x+3-4x^{2}+12x=9$$

Combine 2x and 12x to get 14x.

$$14x+3-4x^{2}=9$$

Subtract 9 from both sides.

$$14x+3-4x^{2}-9=0$$

Subtract 9 from 3 to get -6.

$$14x-6-4x^{2}=0$$

Divide both sides by 2.

$$7x-3-2x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-2x^{2}+7x-3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -2x^{2}+ax+bx-3. To find a and b, set up a system to be solved.

$$a+b=7 ab=-2\left(-3\right)=6$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 6.

$$1,6 2,3$$

Calculate the sum for each pair.

$$1+6=7 2+3=5$$

The solution is the pair that gives sum 7.

$$a=6 b=1$$

Rewrite -2x^{2}+7x-3 as \left(-2x^{2}+6x\right)+\left(x-3\right).

$$\left(-2x^{2}+6x\right)+\left(x-3\right)$$

Factor out 2x in the first and -1 in the second group.

$$2x\left(-x+3\right)-\left(-x+3\right)$$

Factor out common term -x+3 by using distributive property.

$$\left(-x+3\right)\left(2x-1\right)$$

To find equation solutions, solve -x+3=0 and 2x-1=0.

$$x=3 x=\frac{1}{2}$$

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

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

Simplify. The value x=3 does not satisfy the equation because the left and the right hand side have opposite signs.

$$-1=5$$

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

$$2-\sqrt{2\times \frac{1}{2}+3}=2\times \frac{1}{2}-1$$

Simplify. The value x=\frac{1}{2} satisfies the equation.

$$0=0$$

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

$$x=\frac{1}{2}$$