Solve frac{sqrt{2x+10}}{3x-1}-1/2=0

$$\frac { \sqrt { 2 x + 10 } } { 3 x - 1 } - \frac { 1 } { 2 } = 0$$

Solve for x

$$x=3$$

Solution Steps

Subtract -\frac{1}{2} from both sides of the equation.

$$\frac{\sqrt{2x+10}}{3x-1}=\frac{1}{2}$$

Variable x cannot be equal to \frac{1}{3} since division by zero is not defined. Multiply both sides of the equation by 2\left(3x-1\right), the least common multiple of 3x-1,2.

$$2\sqrt{2x+10}=3x-1$$

Square both sides of the equation.

$$\left(2\sqrt{2x+10}\right)^{2}=\left(3x-1\right)^{2}$$

Expand \left(2\sqrt{2x+10}\right)^{2}.

$$2^{2}\left(\sqrt{2x+10}\right)^{2}=\left(3x-1\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$4\left(\sqrt{2x+10}\right)^{2}=\left(3x-1\right)^{2}$$

Calculate \sqrt{2x+10} to the power of 2 and get 2x+10.

$$4\left(2x+10\right)=\left(3x-1\right)^{2}$$

Use the distributive property to multiply 4 by 2x+10.

$$8x+40=\left(3x-1\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-1\right)^{2}.

$$8x+40=9x^{2}-6x+1$$

Subtract 9x^{2} from both sides.

$$8x+40-9x^{2}=-6x+1$$

Add 6x to both sides.

$$8x+40-9x^{2}+6x=1$$

Combine 8x and 6x to get 14x.

$$14x+40-9x^{2}=1$$

Subtract 1 from both sides.

$$14x+40-9x^{2}-1=0$$

Subtract 1 from 40 to get 39.

$$14x+39-9x^{2}=0$$

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

$$-9x^{2}+14x+39=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -9x^{2}+ax+bx+39. To find a and b, set up a system to be solved.

$$a+b=14 ab=-9\times 39=-351$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -351.

$$-1,351 -3,117 -9,39 -13,27$$

Calculate the sum for each pair.

$$-1+351=350 -3+117=114 -9+39=30 -13+27=14$$

The solution is the pair that gives sum 14.

$$a=27 b=-13$$

Rewrite -9x^{2}+14x+39 as \left(-9x^{2}+27x\right)+\left(-13x+39\right).

$$\left(-9x^{2}+27x\right)+\left(-13x+39\right)$$

Factor out 9x in the first and 13 in the second group.

$$9x\left(-x+3\right)+13\left(-x+3\right)$$

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

$$\left(-x+3\right)\left(9x+13\right)$$

To find equation solutions, solve -x+3=0 and 9x+13=0.

$$x=3 x=-\frac{13}{9}$$

Substitute 3 for x in the equation \frac{\sqrt{2x+10}}{3x-1}-\frac{1}{2}=0.

$$\frac{\sqrt{2\times 3+10}}{3\times 3-1}-\frac{1}{2}=0$$

Simplify. The value x=3 satisfies the equation.

$$0=0$$

Substitute -\frac{13}{9} for x in the equation \frac{\sqrt{2x+10}}{3x-1}-\frac{1}{2}=0.

$$\frac{\sqrt{2\left(-\frac{13}{9}\right)+10}}{3\left(-\frac{13}{9}\right)-1}-\frac{1}{2}=0$$

Simplify. The value x=-\frac{13}{9} does not satisfy the equation.

$$-1=0$$

Equation 2\sqrt{2x+10}=3x-1 has a unique solution.

$$x=3$$