Solve sqrt{a^2+1}+a=2

$$\sqrt { a ^ { 2 } + 1 } + a = 2$$

Solve for a

$$a=\frac{3}{4}=0.75$$

Solution Steps

Subtract a from both sides of the equation.

$$\sqrt{a^{2}+1}=2-a$$

Square both sides of the equation.

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

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

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

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

$$a^{2}+1=4-4a+a^{2}$$

Add 4a to both sides.

$$a^{2}+1+4a=4+a^{2}$$

Subtract a^{2} from both sides.

$$a^{2}+1+4a-a^{2}=4$$

Combine a^{2} and -a^{2} to get 0.

$$1+4a=4$$

Subtract 1 from both sides.

$$4a=4-1$$

Subtract 1 from 4 to get 3.

$$4a=3$$

Divide both sides by 4.

$$a=\frac{3}{4}$$

Substitute \frac{3}{4} for a in the equation \sqrt{a^{2}+1}+a=2.

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

Simplify. The value a=\frac{3}{4} satisfies the equation.

$$2=2$$

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

$$a=\frac{3}{4}$$