Step-by-step explanation:
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%5E2, must be non-negative. In other "words":
16-9x%5E2+%3E=+0
This is a quadratic inequality. These can be solved graphically or algebraically.
Graphical solution to 16-9x%5E2+%3E=+0.
Consider the equation
y+=+16-5x%5E2
If know your equations you will recognize this as a parabola (because of the x%5E2 but no y%5E2) which opens downward (because of the minus in front of the x%5E2 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
those x's make y zero or positive
Since y+=++16-4x%5E2, those same x's will make 16-4x%5E2 zero or positive.
We are looking for the x's that make 16-4x%5E2 zero or positive.
Here's a graph to help you see this:
graph%28400%2C+400%2C+-20%2C+20%2C+-20%2C+20%2C+16-4x%5E2%29
We are looking for the x's in that "bump" from the x-axis and up. The x's that make 16-4x%5E2 zero will be where the "bump" intersects the x-axis. We want these x values and everything in between.
To find the x-intercepts we set y to zero (because y is zero on the x-axis) and solve the equation:
16-x%5E2+=+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%2F3+%3C=+x+%3C=+4%2F3
An algebraic solution for 16-9x%5E2+%3E=+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:
Both factors are positive (or zero)
Both factors are negative (or zero)
Algebraically "both factors are positive (or zero)" is expressed as:
4%2B3x+%3E=+0+ and 4-3x+%3E=+0
Algebraically "both factors are negative (or zero)" is expressed as:
4%2B3x+%3C=+0+ and 4-3x+%3C=+0
And finally to "say" that one or the other of these is true would be:
(4%2B3x+%3E=+0+ and 4-3x+%3E=+0) or (4%2B3x+%3C=+0+ and 4-3x+%3C=+0)
We can solve this compound inequality. Solving each of the individual inequalities we get:
(x+%3E=+-4%2F3 and x+%3C=+4%2F3) or (x+%3C=+-4%2F3 and x+%3E=+4%2F3)
(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%2F3+%3C=+x+%3C=+4%2F3 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 at the same time!
So either graphically or algebraically the solution is:
-4%2F3+%3C=+x+%3C=+4%2F3
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+%3E=+0
Most people would start by subtracting 4 from each side:
-3x+%3E=+-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+%3C=+-4%2F3
