Solve a=sqrt{6+a}

$$a = \sqrt { 6 + a }$$

Solve for a

$$a=3$$

Solution Steps

Square both sides of the equation.

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

Calculate \sqrt{6+a} to the power of 2 and get 6+a.

$$a^{2}=6+a$$

Subtract 6 from both sides.

$$a^{2}-6=a$$

Subtract a from both sides.

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

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

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

To solve the equation, factor a^{2}-a-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.

$$a+b=-1 ab=-6$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -6.

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

Calculate the sum for each pair.

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

The solution is the pair that gives sum -1.

$$a=-3 b=2$$

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

$$\left(a-3\right)\left(a+2\right)$$

To find equation solutions, solve a-3=0 and a+2=0.

$$a=3 a=-2$$

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

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

Simplify. The value a=3 satisfies the equation.

$$3=3$$

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

$$-2=\sqrt{6-2}$$

Simplify. The value a=-2 does not satisfy the equation because the left and the right hand side have opposite signs.

$$-2=2$$

Equation a=\sqrt{a+6} has a unique solution.

$$a=3$$