A square substitution preserves an excluded zero

x\le15/(x+2)

Total marks: 5

Solution

(i)

Rational inequalities

Without a graphing calculator, solve $x\le\dfrac{15}{x+2}$ .

Explore this idea

Hint

State $x\ne-2$ . Move $\frac{15}{x+2}$ to the left and form a single fraction.
Factorise the numerator and use a sign chart.

Answer · (i)

x\le-5

Explore the idea

Where is (x + 5)(x − 3)/(x + 2) ≤ 0?

Drag the test value. Closed circles are included, open circles excluded.

\frac{(x+5)(x-3)}{x+2}\le0

Choose an interval to test it. Shading marks the solution intervals.

Factor
x + 5	negative	positive	positive	positive
x − 3	negative	negative	negative	positive
x + 2	negative	negative	positive	positive
product	negative	positive	negative	positive

x + 5	x − 3	x + 2	product
negative	negative	negative	negative
positive	negative	negative	positive
positive	negative	positive	negative
positive	positive	positive	positive

Test value of xx = 0.00

Try an example

(x + 5)+(x − 3)−(x + 2)+⇒ product negative

Is it ≤ 0?

Yes

x is in the solution set

The shaded strips are where the expression satisfies the inequality; they match the green-tinted columns of the chart.

Expression, and where it is ≤ 0
Your test value

Key idea

Zeros of the numerator are included in a ≤ inequality; zeros of the denominator never are.

Ideas to try

Test the three critical values

-5

-2

and

3

. Which are included?

(ii)

Substitution of a square

Hence solve $\dfrac1{x^2}\ge\dfrac{x^2+2}{15}$ .

Explore this idea

Hint

Rearrange to the form of (i) with $t=x^2$ : both $x^2$ and $x^2+2$ are positive.
Substitute $t$ into the earlier answer.
Use $x^2>0$ (since $x\ne0$ ) to cut the answer down, then take square roots.

Answer · (ii)

-\sqrt3\le x<0

Explore the idea

Pull an output interval back through a square

Change the upper bound on t = x². Both square-root branches must be kept, while x = 0 stays excluded.

The original part (i) gives t ≤ −5 or −2 < t ≤ 3. Here t = x² and the original denominator requires x ≠ 0. Reset restores the question’s bound 3.

Upper output bound3

Closed endpoint: the boundary is included unless the original domain excludes it.

Try an example

Required output values

(−∞, -5] ∪ (-2, 3]

Retained inputs

[-1.73205, 0) ∪ (0, 1.73205]

Reached outputs

(0, 3]

Original allowed inputs

(−∞, 0) ∪ (0, ∞)

These exclusions stay in force when a boundary changes. Arrows indicate continuation beyond the drawing window.

The negative output interval is unreachable. The positive part gives two input intervals, with closed outer endpoints. Zero remains excluded in both.

The graph shows a finite window; arrows and interval notation describe the full set. Decimal endpoints are rounded.

A square substitution preserves an excluded zero

Solution

Rational inequalities

One side zero

Single fraction

Sign chart

Where is (x + 5)(x − 3)/(x + 2) ≤ 0?

Substitution of a square

Rearrange

Substitute into the answer to (i)

Keep only possible values

Take square roots