QuestionPart B - Thinking and Investigation Full marks will be given on IT-15 marksl
1. Solve the following inequality using an algebraic method. $\frac{\pi}{x+5}-\frac{1}{x}<2$

Studdy Solution

STEP 1

What is this asking? We need to find all the values of $x$ that make the expression $\frac{1}{x+5} - \frac{1}{x}$ less than 2. Watch out! Don't forget to consider what happens when $x$ or $x+5$ equals zero, and how that affects our inequality!

STEP 2

1. Prepare the inequality
2. Find the critical values
3. Test the intervals

STEP 3

Let's move the 2 to the left side of the inequality to get a zero on the right side.
This helps us compare our expression directly to zero! $\frac{1}{x+5} - \frac{1}{x} - 2 < 0$

STEP 4

To make things easier to work with, let's combine the fractions.
We need a common denominator, which is $x(x+5)$ .
Remember that $x$ and $x+5$ cannot be zero, so $x \neq 0$ and $x \neq -5$ . $\frac{x - (x+5) - 2x(x+5)}{x(x+5)} < 0$

STEP 5

Let's simplify the numerator carefully. $\frac{x - x - 5 - 2x^2 - 10x}{x(x+5)} < 0$ $\frac{-2x^2 - 10x - 5}{x(x+5)} < 0$

STEP 6

Multiplying both sides by -1 flips the inequality sign!
Don't forget this crucial step! $\frac{2x^2 + 10x + 5}{x(x+5)} > 0$

STEP 7

Let's find when the numerator is equal to zero using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here, $a = \textbf{2}$ , $b = \textbf{10}$ , and $c = \textbf{5}$ . $x = \frac{-10 \pm \sqrt{100 - 4 \cdot 2 \cdot 5}}{2 \cdot 2}$ $x = \frac{-10 \pm \sqrt{60}}{4}$ $x = \frac{-10 \pm 2\sqrt{15}}{4}$ $x = \frac{-5 \pm \sqrt{15}}{2}$ So, $x_1 = \frac{-5 - \sqrt{15}}{2} \approx \textbf{-4.436}$ and $x_2 = \frac{-5 + \sqrt{15}}{2} \approx \textbf{-0.564}$ .

STEP 8

The denominator is zero when $x = \textbf{0}$ or $x = \textbf{-5}$ .
These are also critical values.

STEP 9

Our critical values are approximately $-5$ , $-4.436$ , $-0.564$ , and $0$ .
These divide the number line into five intervals: $(-\infty, -5)$ , $(-5, -4.436)$ , $(-4.436, -0.564)$ , $(-0.564, 0)$ , and $(0, \infty)$ .

STEP 10

We need to test a value within each interval to see if the inequality holds true.
If it's true for one value in the interval, it's true for all values in that interval!
* Interval 1: $x = -6$ : The expression becomes positive. * Interval 2: $x = -4.5$ : The expression becomes negative. * Interval 3: $x = -1$ : The expression becomes positive. * Interval 4: $x = -0.1$ : The expression becomes negative. * Interval 5: $x = 1$ : The expression becomes positive.

STEP 11

The inequality is true when $x$ is in the intervals $(-\infty, -5)$ , $(-4.436, -0.564)$ , and $(0, \infty)$ .
More precisely, $x < -5$ , or $\frac{-5 - \sqrt{15}}{2} < x < \frac{-5 + \sqrt{15}}{2}$ , or $x > 0$ .

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