Solution

Can we find a quadratic inequality for each region of the Venn diagram?

The regions are defined as follows.

A: The solution set is a subset of \(x≤1\).

B: The solutions are given by \(a≤x≤b\) where \(a\) and \(b\) are real numbers.

C: The inequality is satisfied by \(x=4\), e.g. \(x=4\) satisfies the inequality \(x≥2\).

We will solve the given quadratic inequalities so they can be placed in the Venn diagram. We will start with the more straightforward ones.

\(x^2≤9\)

\(11x≥2x^2\)

Does it matter whether we move the terms to the left or the right hand side of the inequality? What would change if we solved \(11x - 2x^2 ≥ 0\) instead?

\(6x^2-1≥5x\)

\(3x^2≥21x-30\)

Why can we divide both sides by \(3\)? How does the graph of \(y = 3x^2 - 21x + 30\) compare to the graph of \(y = x^2 - 7x +10\)?

\(x^2≤-x\)

\(-2x^2≤x-6\)

\(x^2+3≥2\)

We didn’t try to solve this algebraically because of what we know about \(x^2\). What would happen if we did try to solve it algebraically?

\(x^2≤x-2\)

What other graph might we have sketched to see there are no solutions?


Here are the given inequalities presented in the Venn Diagram.

Venn diagram showing where each inequality lies with empty regions A and C, and A, B and C.

It is impossible for a number to be less than 1 and equal to 4 simultaneously. Therefore it is impossible to fill these regions.

One of the ways for the solution set to be a subset of \(x≤1\) is if it is bounded by two values, e.g. \(-3≤x≤0\), but this would mean the solution set also satisfies condition B.

Another would be if the solution set only has one bound, e.g. \(x ≤-4\), but this is impossible for a quadratic inequality.

However, we could consider a case whose solution set is a point, e.g. \((x+1)^2≤0\) has the solution set \(x=-1\) which satisfies condition A only.

There is some discussion about this solution, as this can also be written as \(-1≤x≤-1\) which then means it also satisfies condition B.

You might have suggested a quadratic inequality with a solution set such as \(-3<x<0\) (using different inequality signs) and this would indeed satisfy only condition A. However we chose to only use the signs \(≤\) and \(≥\).