Solve x=sqrt{3-x/2}?

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

Solve for x

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

Solution Steps

Square both sides of the equation.

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

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

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

Multiply both sides of the equation by 2.

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

Subtract 6 from both sides.

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

Add x to both sides.

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

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

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

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

$$a+b=1 ab=2\left(-6\right)=-12$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -12.

$$-1,12 -2,6 -3,4$$

Calculate the sum for each pair.

$$-1+12=11 -2+6=4 -3+4=1$$

The solution is the pair that gives sum 1.

$$a=-3 b=4$$

Rewrite 2x^{2}+x-6 as \left(2x^{2}-3x\right)+\left(4x-6\right).

$$\left(2x^{2}-3x\right)+\left(4x-6\right)$$

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

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

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

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

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

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

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

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

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

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

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

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

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

$$-2=2$$

Equation x=\sqrt{-\frac{x}{2}+3} has a unique solution.

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