ir2pid wrote:

(x+5)(3−x)>0. This equation holds true for −5>x>3.

Without applying values for x and testing, what is the simplest way of getting the range of x as −5>x>3

Dear

ir2pidIn questions like the one you posted,

the Wavy Line Method is an elegant and quick way to find the range of x.

Here is what you do in the Wavy Line Method:

1. Express the given expression in

standard factorized form, that is in the form:

\((x-a)^{p}(x-b)^{q}\) < or > 0

The inequality you presented above is already in factorized form.

But is it in Standard form? Not yet. [because one of the factors is (3-x). We need to convert it to the form (x-3)]

Let's multiply both sides of the inequality with -1. Note that multiplying both sides of an inequality with a negative number changes the sign of inequality. So, we get:

(x+5)(x-3) < 0

2.

Plot the Zero Points on the number line. These are the points for which the factorized expression becomes equal to zero.

(x+5)(x-3) becomes equal to zero for x = -5 and x = 3. So, these are the zero points. Plot them on the number line.

3. Now, starting from the

TOP RIGHT CORNER of the number line,

draw a Wavy Line that passes through these zero points.

The parts where this wavy line is

ABOVE the number line are the parts where the given expression will be

positive.

The parts where this wavy line is

BELOW the number line are the parts where the given expression will be

negative.

That's it!

From the Wavy Line we've drawn for your inequality, you can easily see that that the expression (x+5)(x-3) will be negative for -5 < x < 3

Practice the Wavy Line Method for one or two more questions and am sure you'll start finding questions like this one to be pretty easy.

Hope this was useful!

Japinder

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