Solve sqrt{15-x}+sqrt{3-x}=6

$$\sqrt{ 15-x } + \sqrt{ 3-x } =6$$

Solve for x

$$x=-1$$

Solution Steps

Subtract \sqrt{3-x} from both sides of the equation.

$$\sqrt{15-x}=6-\sqrt{3-x}$$

Square both sides of the equation.

$$\left(\sqrt{15-x}\right)^{2}=\left(6-\sqrt{3-x}\right)^{2}$$

Calculate \sqrt{15-x} to the power of 2 and get 15-x.

$$15-x=\left(6-\sqrt{3-x}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(6-\sqrt{3-x}\right)^{2}.

$$15-x=36-12\sqrt{3-x}+\left(\sqrt{3-x}\right)^{2}$$

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

$$15-x=36-12\sqrt{3-x}+3-x$$

Add 36 and 3 to get 39.

$$15-x=39-12\sqrt{3-x}-x$$

Add 12\sqrt{3-x} to both sides.

$$15-x+12\sqrt{3-x}=39-x$$

Add x to both sides.

$$15-x+12\sqrt{3-x}+x=39$$

Combine -x and x to get 0.

$$15+12\sqrt{3-x}=39$$

Subtract 15 from both sides.

$$12\sqrt{3-x}=39-15$$

Subtract 15 from 39 to get 24.

$$12\sqrt{3-x}=24$$

Divide both sides by 12.

$$\sqrt{3-x}=\frac{24}{12}$$

Divide 24 by 12 to get 2.

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

Square both sides of the equation.

$$-x+3=4$$

Subtract 3 from both sides of the equation.

$$-x+3-3=4-3$$

Subtracting 3 from itself leaves 0.

$$-x=4-3$$

Subtract 3 from 4.

$$-x=1$$

Divide both sides by -1.

$$\frac{-x}{-1}=\frac{1}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x=\frac{1}{-1}$$

Divide 1 by -1.

$$x=-1$$

Substitute -1 for x in the equation \sqrt{15-x}+\sqrt{3-x}=6.

$$\sqrt{15-\left(-1\right)}+\sqrt{3-\left(-1\right)}=6$$

Simplify. The value x=-1 satisfies the equation.

$$6=6$$

Equation \sqrt{15-x}=-\sqrt{3-x}+6 has a unique solution.

$$x=-1$$