Solve sqrt{2x-1}-3x+2=0

$$\sqrt{ 2x-1 } -3x+2=0$$

Solve for x

$$x=1$$

Solution Steps

Subtract -3x+2 from both sides of the equation.

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

To find the opposite of -3x+2, find the opposite of each term.

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

The opposite of -3x is 3x.

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

Square both sides of the equation.

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

Calculate \sqrt{2x-1} to the power of 2 and get 2x-1.

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

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

$$2x-1=9x^{2}-12x+4$$

Subtract 9x^{2} from both sides.

$$2x-1-9x^{2}=-12x+4$$

Add 12x to both sides.

$$2x-1-9x^{2}+12x=4$$

Combine 2x and 12x to get 14x.

$$14x-1-9x^{2}=4$$

Subtract 4 from both sides.

$$14x-1-9x^{2}-4=0$$

Subtract 4 from -1 to get -5.

$$14x-5-9x^{2}=0$$

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

$$-9x^{2}+14x-5=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -9x^{2}+ax+bx-5. To find a and b, set up a system to be solved.

$$a+b=14 ab=-9\left(-5\right)=45$$

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 45.

$$1,45 3,15 5,9$$

Calculate the sum for each pair.

$$1+45=46 3+15=18 5+9=14$$

The solution is the pair that gives sum 14.

$$a=9 b=5$$

Rewrite -9x^{2}+14x-5 as \left(-9x^{2}+9x\right)+\left(5x-5\right).

$$\left(-9x^{2}+9x\right)+\left(5x-5\right)$$

Factor out 9x in the first and -5 in the second group.

$$9x\left(-x+1\right)-5\left(-x+1\right)$$

Factor out common term -x+1 by using distributive property.

$$\left(-x+1\right)\left(9x-5\right)$$

To find equation solutions, solve -x+1=0 and 9x-5=0.

$$x=1 x=\frac{5}{9}$$

Substitute 1 for x in the equation \sqrt{2x-1}-3x+2=0.

$$\sqrt{2\times 1-1}-3+2=0$$

Simplify. The value x=1 satisfies the equation.

$$0=0$$

Substitute \frac{5}{9} for x in the equation \sqrt{2x-1}-3x+2=0.

$$\sqrt{2\times \frac{5}{9}-1}-3\times \frac{5}{9}+2=0$$

Simplify. The value x=\frac{5}{9} does not satisfy the equation.

$$\frac{2}{3}=0$$

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

$$x=1$$