Solve sqrt{3a+1}+3=a

$$\sqrt { 3 a + 1 } + 3 = a$$

Solve for a

$$a=8$$

Solution Steps

Subtract 3 from both sides of the equation.

$$\sqrt{3a+1}=a-3$$

Square both sides of the equation.

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

Calculate \sqrt{3a+1} to the power of 2 and get 3a+1.

$$3a+1=\left(a-3\right)^{2}$$

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

$$3a+1=a^{2}-6a+9$$

Subtract a^{2} from both sides.

$$3a+1-a^{2}=-6a+9$$

Add 6a to both sides.

$$3a+1-a^{2}+6a=9$$

Combine 3a and 6a to get 9a.

$$9a+1-a^{2}=9$$

Subtract 9 from both sides.

$$9a+1-a^{2}-9=0$$

Subtract 9 from 1 to get -8.

$$9a-8-a^{2}=0$$

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

$$-a^{2}+9a-8=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -a^{2}+aa+ba-8. To find a and b, set up a system to be solved.

$$a+b=9 ab=-\left(-8\right)=8$$

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 8.

$$1,8 2,4$$

Calculate the sum for each pair.

$$1+8=9 2+4=6$$

The solution is the pair that gives sum 9.

$$a=8 b=1$$

Rewrite -a^{2}+9a-8 as \left(-a^{2}+8a\right)+\left(a-8\right).

$$\left(-a^{2}+8a\right)+\left(a-8\right)$$

Factor out -a in -a^{2}+8a.

$$-a\left(a-8\right)+a-8$$

Factor out common term a-8 by using distributive property.

$$\left(a-8\right)\left(-a+1\right)$$

To find equation solutions, solve a-8=0 and -a+1=0.

$$a=8 a=1$$

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

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

Simplify. The value a=8 satisfies the equation.

$$8=8$$

Substitute 1 for a in the equation \sqrt{3a+1}+3=a.

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

Simplify. The value a=1 does not satisfy the equation.

$$5=1$$

Equation \sqrt{3a+1}=a-3 has a unique solution.

$$a=8$$