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

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

Solve for x

$$x=3$$

Solution Steps

Square both sides of the equation.

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

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

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

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

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

Subtract 3 from both sides of the equation.

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

Square both sides of the equation.

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

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

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

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

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

Subtract x^{2} from both sides.

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

Add 6x to both sides.

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

Combine -x and 6x to get 5x.

$$3+5x-x^{2}=9$$

Subtract 9 from both sides.

$$3+5x-x^{2}-9=0$$

Subtract 9 from 3 to get -6.

$$-6+5x-x^{2}=0$$

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

$$-x^{2}+5x-6=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-6. To find a and b, set up a system to be solved.

$$a+b=5 ab=-\left(-6\right)=6$$

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 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 5.

$$a=3 b=2$$

Rewrite -x^{2}+5x-6 as \left(-x^{2}+3x\right)+\left(2x-6\right).

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

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

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

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

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

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

$$x=3 x=2$$

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

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

Simplify. The value x=3 satisfies the equation.

$$3^{\frac{1}{2}}=3^{\frac{1}{2}}$$

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

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

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

$$2=2^{\frac{1}{2}}$$

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

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

Simplify. The value x=3 satisfies the equation.

$$3^{\frac{1}{2}}=3^{\frac{1}{2}}$$

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

$$x=3$$