Solve 2x+sqrt{x+1}=8

$$2 x + \sqrt { x + 1 } = 8$$

Solve for x

$$x=3$$

Solution Steps

Subtract 2x from both sides of the equation.

$$\sqrt{x+1}=8-2x$$

Square both sides of the equation.

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

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

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(8-2x\right)^{2}.

$$x+1=64-32x+4x^{2}$$

Subtract 64 from both sides.

$$x+1-64=-32x+4x^{2}$$

Subtract 64 from 1 to get -63.

$$x-63=-32x+4x^{2}$$

Add 32x to both sides.

$$x-63+32x=4x^{2}$$

Combine x and 32x to get 33x.

$$33x-63=4x^{2}$$

Subtract 4x^{2} from both sides.

$$33x-63-4x^{2}=0$$

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

$$-4x^{2}+33x-63=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -4x^{2}+ax+bx-63. To find a and b, set up a system to be solved.

$$a+b=33 ab=-4\left(-63\right)=252$$

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 252.

$$1,252 2,126 3,84 4,63 6,42 7,36 9,28 12,21 14,18$$

Calculate the sum for each pair.

$$1+252=253 2+126=128 3+84=87 4+63=67 6+42=48 7+36=43 9+28=37 12+21=33 14+18=32$$

The solution is the pair that gives sum 33.

$$a=21 b=12$$

Rewrite -4x^{2}+33x-63 as \left(-4x^{2}+21x\right)+\left(12x-63\right).

$$\left(-4x^{2}+21x\right)+\left(12x-63\right)$$

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

$$-x\left(4x-21\right)+3\left(4x-21\right)$$

Factor out common term 4x-21 by using distributive property.

$$\left(4x-21\right)\left(-x+3\right)$$

To find equation solutions, solve 4x-21=0 and -x+3=0.

$$x=\frac{21}{4} x=3$$

Substitute \frac{21}{4} for x in the equation 2x+\sqrt{x+1}=8.

$$2\times \frac{21}{4}+\sqrt{\frac{21}{4}+1}=8$$

Simplify. The value x=\frac{21}{4} does not satisfy the equation.

$$13=8$$

Substitute 3 for x in the equation 2x+\sqrt{x+1}=8.

$$2\times 3+\sqrt{3+1}=8$$

Simplify. The value x=3 satisfies the equation.

$$8=8$$

Equation \sqrt{x+1}=8-2x has a unique solution.

$$x=3$$