Solve sqrt{2a-3}+3=a

$$\sqrt { 2 a - 3 } + 3 = a$$

Solve for a

$$a=6$$

Solution Steps

Subtract 3 from both sides of the equation.

$$\sqrt{2a-3}=a-3$$

Square both sides of the equation.

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

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

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

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

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

Subtract a^{2} from both sides.

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

Add 6a to both sides.

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

Combine 2a and 6a to get 8a.

$$8a-3-a^{2}=9$$

Subtract 9 from both sides.

$$8a-3-a^{2}-9=0$$

Subtract 9 from -3 to get -12.

$$8a-12-a^{2}=0$$

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

$$-a^{2}+8a-12=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -a^{2}+aa+ba-12. To find a and b, set up a system to be solved.

$$a+b=8 ab=-\left(-12\right)=12$$

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 12.

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

Calculate the sum for each pair.

$$1+12=13 2+6=8 3+4=7$$

The solution is the pair that gives sum 8.

$$a=6 b=2$$

Rewrite -a^{2}+8a-12 as \left(-a^{2}+6a\right)+\left(2a-12\right).

$$\left(-a^{2}+6a\right)+\left(2a-12\right)$$

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

$$-a\left(a-6\right)+2\left(a-6\right)$$

Factor out common term a-6 by using distributive property.

$$\left(a-6\right)\left(-a+2\right)$$

To find equation solutions, solve a-6=0 and -a+2=0.

$$a=6 a=2$$

Substitute 6 for a in the equation \sqrt{2a-3}+3=a.

$$\sqrt{2\times 6-3}+3=6$$

Simplify. The value a=6 satisfies the equation.

$$6=6$$

Substitute 2 for a in the equation \sqrt{2a-3}+3=a.

$$\sqrt{2\times 2-3}+3=2$$

Simplify. The value a=2 does not satisfy the equation.

$$4=2$$

Equation \sqrt{2a-3}=a-3 has a unique solution.

$$a=6$$