Solve sqrt{4x+21}-3=2x

$$\sqrt{ 4x+21 } -3 = 2x$$

Solve for x

$$x=1$$

Solution Steps

Subtract -3 from both sides of the equation.

$$\sqrt{4x+21}=2x+3$$

Square both sides of the equation.

$$\left(\sqrt{4x+21}\right)^{2}=\left(2x+3\right)^{2}$$

Calculate \sqrt{4x+21} to the power of 2 and get 4x+21.

$$4x+21=\left(2x+3\right)^{2}$$

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

$$4x+21=4x^{2}+12x+9$$

Subtract 4x^{2} from both sides.

$$4x+21-4x^{2}=12x+9$$

Subtract 12x from both sides.

$$4x+21-4x^{2}-12x=9$$

Combine 4x and -12x to get -8x.

$$-8x+21-4x^{2}=9$$

Subtract 9 from both sides.

$$-8x+21-4x^{2}-9=0$$

Subtract 9 from 21 to get 12.

$$-8x+12-4x^{2}=0$$

Divide both sides by 4.

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

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

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

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

$$a+b=-2 ab=-3=-3$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. The only such pair is the system solution.

$$a=1 b=-3$$

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

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

Factor out x in the first and 3 in the second group.

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

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

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

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

$$x=1 x=-3$$

Substitute 1 for x in the equation \sqrt{4x+21}-3=2x.

$$\sqrt{4\times 1+21}-3=2\times 1$$

Simplify. The value x=1 satisfies the equation.

$$2=2$$

Substitute -3 for x in the equation \sqrt{4x+21}-3=2x.

$$\sqrt{4\left(-3\right)+21}-3=2\left(-3\right)$$

Simplify. The value x=-3 does not satisfy the equation.

$$0=-6$$

Equation \sqrt{4x+21}=2x+3 has a unique solution.

$$x=1$$