Solve sqrt{3}sqrt{9-x^2}=3+x

$$\sqrt { 3 } \sqrt { 9 - x ^ { 2 } } = 3 + x$$

Solve for x

$$x = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

$$x=-3$$

Solution Steps

Square both sides of the equation.

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

Expand \left(\sqrt{3}\sqrt{9-x^{2}}\right)^{2}.

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

The square of \sqrt{3} is 3.

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

Calculate \sqrt{9-x^{2}} to the power of 2 and get 9-x^{2}.

$$3\left(9-x^{2}\right)=\left(3+x\right)^{2}$$

Use the distributive property to multiply 3 by 9-x^{2}.

$$27-3x^{2}=\left(3+x\right)^{2}$$

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

$$27-3x^{2}=9+6x+x^{2}$$

Subtract 9 from both sides.

$$27-3x^{2}-9=6x+x^{2}$$

Subtract 9 from 27 to get 18.

$$18-3x^{2}=6x+x^{2}$$

Subtract 6x from both sides.

$$18-3x^{2}-6x=x^{2}$$

Subtract x^{2} from both sides.

$$18-3x^{2}-6x-x^{2}=0$$

Combine -3x^{2} and -x^{2} to get -4x^{2}.

$$18-4x^{2}-6x=0$$

Divide both sides by 2.

$$9-2x^{2}-3x=0$$

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

$$-2x^{2}-3x+9=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -2x^{2}+ax+bx+9. To find a and b, set up a system to be solved.

$$a+b=-3 ab=-2\times 9=-18$$

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 -18.

$$1,-18 2,-9 3,-6$$

Calculate the sum for each pair.

$$1-18=-17 2-9=-7 3-6=-3$$

The solution is the pair that gives sum -3.

$$a=3 b=-6$$

Rewrite -2x^{2}-3x+9 as \left(-2x^{2}+3x\right)+\left(-6x+9\right).

$$\left(-2x^{2}+3x\right)+\left(-6x+9\right)$$

Factor out -x in the first and -3 in the second group.

$$-x\left(2x-3\right)-3\left(2x-3\right)$$

Factor out common term 2x-3 by using distributive property.

$$\left(2x-3\right)\left(-x-3\right)$$

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

$$x=\frac{3}{2} x=-3$$

Substitute \frac{3}{2} for x in the equation \sqrt{3}\sqrt{9-x^{2}}=3+x.

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

Simplify. The value x=\frac{3}{2} satisfies the equation.

$$\frac{9}{2}=\frac{9}{2}$$

Substitute -3 for x in the equation \sqrt{3}\sqrt{9-x^{2}}=3+x.

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

Simplify. The value x=-3 satisfies the equation.

$$0=0$$

List all solutions of \sqrt{3}\sqrt{9-x^{2}}=x+3.

$$x=\frac{3}{2} x=-3$$