Solve 3-x=sqrt{3+x}

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

Solve for x

$$x=1$$

Solution Steps

Square both sides of the equation.

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

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

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

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

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

Subtract 3 from both sides.

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

Subtract 3 from 9 to get 6.

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

Subtract x from both sides.

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

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

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

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

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

To solve the equation, factor x^{2}-7x+6 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+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(x+a\right)\left(x+b\right) using the obtained values.

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

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

$$x=6 x=1$$

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

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

Simplify. The value x=6 does not satisfy the equation because the left and the right hand side have opposite signs.

$$-3=3$$

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

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

Simplify. The value x=1 satisfies the equation.

$$2=2$$

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

$$x=1$$