Solve (x)=sqrt{2x+3/2x+3}

$$( x ) = \sqrt { \frac { 2 x + 3 } { 2 x + 3 } }$$

Solve for x

$$x=1$$

Solution Steps

Square both sides of the equation.

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

Cancel out 2x+3 in both numerator and denominator.

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

The square of \sqrt{1} is 1.

$$x^{2}=1$$

Subtract 1 from both sides.

$$x^{2}-1=0$$

Consider x^{2}-1. Rewrite x^{2}-1 as x^{2}-1^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(x-1\right)\left(x+1\right)=0$$

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

$$x=1 x=-1$$

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

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

Simplify. The value x=1 satisfies the equation.

$$1=1$$

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

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

Simplify. The value x=-1 does not satisfy the equation because the left and the right hand side have opposite signs.

$$-1=1$$

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

$$x=1$$