Replace the input by its magnitude

(a) Solve (x2−6)/(2x)>1(x^2-6)/(2x)>1 exactly. marks Hence solve (x2−6)/(2∣x∣)≥1(x^2-6)/(2|x|)\ge1. marks

(b) Prove algebraically that x2−4x+9>0x^2-4x+9>0 for all real x. Hence solve (x3+2x2+x+14)/(x2+5)>x+1(x^3+2x^2+x+14)/(x^2+5)>x+1.

[2008/RJC/J1/T3 CT/9]

Total marks: 5

Solution

(a)i

Rational inequalities

Solve x2−62x>1\dfrac{x^2-6}{2x}>1 exactly.

Explore this idea

Hint

  1. Move 1 to the left and combine: do not cross-multiply by 2x2x.

  2. The numerator does not factorise over the integers: complete the square.

  3. Use a sign chart.

Answer · (a)i

1−7<x<01-\sqrt7<x<0 or x>1+7x>1+\sqrt7.

Explore the idea

Where is (x² − 2x − 6)/(2x) positive?

Three critical values: the two roots of the numerator and the zero of the denominator.

(x−1−7)(x−1+7)2x>0\frac{\left(x-1-\sqrt7\right)\left(x-1+\sqrt7\right)}{2x}>0

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

Intervalx − (1 − √7)x − (1 + √7)xproduct
negativenegativenegativenegative
positivenegativenegativepositive
positivenegativepositivenegative
positivepositivepositivepositive
Test value of xx = 1.00
Try an example

(x − (1 − √7))+(x − (1 + √7))−(x)+⇒ product negative

Is it > 0?
No
x is not a solution

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

  • Expression, and the x-values where it is positive
  • Your test value
Key idea

Complete the square when a quadratic's roots are not integers; then the sign chart is the same as for any other fraction.

Ideas to try
  1. Which interval contains x=−1x=-1? Test it.
  2. Why is x=0x=0 never a solution?
(a)ii

Absolute value substitution

Hence solve x2−62∣x∣≥1\dfrac{x^2-6}{2|x|}\ge1.

Explore this idea

Hint

  1. Compare with (i): x2=∣x∣2x^2=|x|^2, so the expression is ∣x∣2−62∣x∣\frac{|x|^2-6}{2|x|}: part (i) with xx replaced by ∣x∣|x|.

  2. The inequality is now non-strict, so the endpoints count.

  3. Translate ∣x∣>…|x|>\dots back into xx.

Answer · (a)ii

x≤−1−7x\le-1-\sqrt7 or x≥1+7x\ge1+\sqrt7.

Explore the idea

Replacing x by |x| mirrors the graph

The graph of (x² − 6)/(2|x|) is symmetric about the y-axis. The line y = 1 cuts it at x = ±(1 + √7).

Switch on a feature to add it to the sketch and see how it is found. A full-mark sketch shows all of them.

  • Version with |x|, symmetric about the y-axis
  • Original without the absolute value, for comparison
  • Y = 1
Ideas to try
  1. Compare the two curves for x>0x>0 and x<0x<0.
  2. Where is the dashed curve above y=1y=1 for negative xx?
(b)i

Completing the square

Prove algebraically that x2−4x+9>0x^2-4x+9>0 for all real xx.

Hint

  1. Complete the square.

  2. Compare with 0.

Answer · (b)i

x2−4x+9=(x−2)2+5>0x^2-4x+9=(x-2)^2+5>0 for all real xx.

(b)ii

Using a positivity result

Hence solve x3+2x2+x+14x2+5>x+1\dfrac{x^3+2x^2+x+14}{x^2+5}>x+1.

Explore this idea

Hint

  1. Move x+1x+1 to the left and combine over x2+5x^2+5.

  2. x2+5>0x^2+5>0 always, so the sign is the sign of the numerator.

  3. Compare the numerator with (b)i.

Answer · (b)ii

x∈Rx\in\mathbb R.

Explore the idea

The curve never meets y = x + 1

The difference between the curve and the line is (x² − 4x + 9)/(x² + 5), which is positive everywhere.

Switch on a feature to add it to the sketch and see how it is found. A full-mark sketch shows all of them.

  • Left-hand side
  • Y = x + 1
Ideas to try
Is the teal curve ever below the amber line?