Question 411562
{{{sqrt(16-9x^2)}}}
In order for an even-numbered root (square root, 4th root, 6th root, etc.) to be a real nnumber, the radicand (the expression inside the radical) must be non-negative (positive or zero). So in order for your square root to be real, the radicand, {{{16-9x^2}}}, must be non-negative. In other "words":
{{{16-9x^2 >= 0}}}
This is a quadratic inequality. These can be solved graphically or algebraically.<br>
Graphical solution to {{{16-9x^2 >= 0}}}.
Consider the equation
{{{y = 16-5x^2}}}
If know your equations you will recognize this as a parabola (because of the {{{x^2}}} but no {{{y^2}}}) which opens downward (because of the minus in front of the {{{x^2}}} term). The part of this parabola from the x-axis and up (if any) will have y coordinates that are zero or positive. The x values for this part of the parabola represent the solution to your problem because<ul><li>those x's make y zero or positive</li><li>Since {{{y =  16-4x^2}}}, those same x's will make {{{16-4x^2}}} zero or positive.</li><li>We are looking for the x's that make {{{16-4x^2}}} zero or positive.</li></ul>
Here's a graph to help you see this:
{{{graph(400, 400, -20, 20, -20, 20, 16-4x^2)}}}
We are looking for the x's in that "bump" from the x-axis and up. The x's that make {{{16-4x^2}}} zero will be where the "bump" intersects the x-axis. We want these x values and everything in between.<br>
To find the x-intercepts we set y to zero (because y is zero on the x-axis) and solve the equation:
{{{16-x^2 = 0}}}
Factoring we get:
(4+3x)(4-3x) = 0
From the Zero Product Property we know that one of these factors must be zero:
4+3x = 0 or 4-3x=0
Solving these we get:
x = -4/3 or x = 4/3
So the solution to your problem is all x's between -4/3 and 4/3, inclusive. In algebraic notation this is:
{{{-4/3 <= x <= 4/3}}}<br>
An algebraic solution for {{{16-9x^2 >= 0}}}
First we factor:
(4+3x)(4-3x) >= 0
We have a product that is greater than or equal to zero. With a little thought we can figure out there are two ways this could happen:<ul><li>Both factors are positive (or zero)</li><li>Both factors are negative (or zero)</li></ul>
Algebraically "both factors are positive (or zero)" is expressed as:
{{{4+3x >= 0 }}} and {{{4-3x >= 0}}}
Algebraically "both factors are negative (or zero)" is expressed as:
{{{4+3x <= 0 }}} and {{{4-3x <= 0}}}
And finally to "say" that one or the other of these is true would be:
({{{4+3x >= 0 }}} and {{{4-3x >= 0}}}) or ({{{4+3x <= 0 }}} and {{{4-3x <= 0}}})
We can solve this compound inequality. Solving each of the individual inequalities we get:
({{{x >= -4/3}}} and {{{x <= 4/3}}}) or ({{{x <= -4/3}}} and {{{x >= 4/3}}})
(Note: If you cannot see how the 2nd and 4th inequalities work out the way they do, see below.)
The first pair can be expressed as {{{-4/3 <= x <= 4/3}}} which is the same answer we got in the graphical solution. The second pair has NO solution because it is impossible for x to be less than or equal to -4/3 and greater than or equal to 4/3 <i>at the same time!</i><br>
So either graphically or algebraically the solution is:
{{{-4/3 <= x <= 4/3}}}<br>
If you had trouble solving the 4-3x inequalities....
Because of the minus in front of the x these are a little tricky to solve. For example:
{{{4-3x >= 0}}}
Most people would start by subtracting 4 from each side:
{{{-3x >= -4}}}
and then dividing by -3. But here is the "trick". We have to remember the special rule that applies whenever you multiply or divide both sides of an inequality by any negative number: You must reverse the inequality symbol! So when we divide by -3 the inequality gets reversed:
{{{x <= -4/3}}}