Solve x+sqrt{4x+1}=5

$$x + \sqrt { 4 x + 1 } = 5$$

Solve for x

$$x=2$$

Solution Steps

Subtract x from both sides of the equation.

$$\sqrt{4x+1}=5-x$$

Square both sides of the equation.

$$\left(\sqrt{4x+1}\right)^{2}=\left(5-x\right)^{2}$$

Calculate \sqrt{4x+1} to the power of 2 and get 4x+1.

$$4x+1=\left(5-x\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5-x\right)^{2}.

$$4x+1=25-10x+x^{2}$$

Subtract 25 from both sides.

$$4x+1-25=-10x+x^{2}$$

Subtract 25 from 1 to get -24.

$$4x-24=-10x+x^{2}$$

Add 10x to both sides.

$$4x-24+10x=x^{2}$$

Combine 4x and 10x to get 14x.

$$14x-24=x^{2}$$

Subtract x^{2} from both sides.

$$14x-24-x^{2}=0$$

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

$$-x^{2}+14x-24=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-24. To find a and b, set up a system to be solved.

$$a+b=14 ab=-\left(-24\right)=24$$

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 24.

$$1,24 2,12 3,8 4,6$$

Calculate the sum for each pair.

$$1+24=25 2+12=14 3+8=11 4+6=10$$

The solution is the pair that gives sum 14.

$$a=12 b=2$$

Rewrite -x^{2}+14x-24 as \left(-x^{2}+12x\right)+\left(2x-24\right).

$$\left(-x^{2}+12x\right)+\left(2x-24\right)$$

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

$$-x\left(x-12\right)+2\left(x-12\right)$$

Factor out common term x-12 by using distributive property.

$$\left(x-12\right)\left(-x+2\right)$$

To find equation solutions, solve x-12=0 and -x+2=0.

$$x=12 x=2$$

Substitute 12 for x in the equation x+\sqrt{4x+1}=5.

$$12+\sqrt{4\times 12+1}=5$$

Simplify. The value x=12 does not satisfy the equation.

$$19=5$$

Substitute 2 for x in the equation x+\sqrt{4x+1}=5.

$$2+\sqrt{4\times 2+1}=5$$

Simplify. The value x=2 satisfies the equation.

$$5=5$$

Equation \sqrt{4x+1}=5-x has a unique solution.

$$x=2$$