Solve a-3=sqrt{a+3}

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

Solve for a

$$a=6$$

Solution Steps

Square both sides of the equation.

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

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

$$a^{2}-6a+9=\left(\sqrt{a+3}\right)^{2}$$

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

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

Subtract a from both sides.

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

Combine -6a and -a to get -7a.

$$a^{2}-7a+9=3$$

Subtract 3 from both sides.

$$a^{2}-7a+9-3=0$$

Subtract 3 from 9 to get 6.

$$a^{2}-7a+6=0$$

To solve the equation, factor a^{2}-7a+6 using formula a^{2}+\left(a+b\right)a+ab=\left(a+a\right)\left(a+b\right). To find a and b, set up a system to be solved.

$$a+b=-7 ab=6$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 6.

$$-1,-6 -2,-3$$

Calculate the sum for each pair.

$$-1-6=-7 -2-3=-5$$

The solution is the pair that gives sum -7.

$$a=-6 b=-1$$

Rewrite factored expression \left(a+a\right)\left(a+b\right) using the obtained values.

$$\left(a-6\right)\left(a-1\right)$$

To find equation solutions, solve a-6=0 and a-1=0.

$$a=6 a=1$$

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

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

Simplify. The value a=6 satisfies the equation.

$$3=3$$

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

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

Simplify. The value a=1 does not satisfy the equation because the left and the right hand side have opposite signs.

$$-2=2$$

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

$$a=6$$