A22 - Quadratic inequalities, set notation and regions (Higher)

Reading quadratic signs between roots (Higher)

A game allows scores $(x,y)$ when $x\geqslant1$, $y\geqslant0$ and $x+y<6$. Does $(2,3)$ qualify? What about $(2,4)$? We will describe the whole allowed region, including which edges count. First, use one-variable inequalities to practise choosing whole ranges rather than just boundary points.

For $q(x)=x^2-x-6=(x+2)(x-3)$, the roots are $-2$ and $3$. At $x=0$, the product is $2\times(-3)=-6$, so the expression is negative between the roots. To the left, both factors are negative, so their product is positive. To the right, both are positive. A sign sketch is a picture of this product reasoning.

[DIAGRAM: asset_name: A quadratic's positive and negative intervals; asset_slug: aqa8300_a30_quadratic_signs; caption: The roots split the number line into intervals where the quadratic keeps the same sign; description: An upward-opening curve y equals x squared minus x minus 6 crossing at minus 2 and 3. The curve is clay below the axis and blue above. A sign strip beneath shows positive to the left of minus 2, negative between minus 2 and 3, positive to the right of 3, and zero at the roots.]

The equation $q(x)=0$ finds only the boundaries. For an inequality, choose the intervals with the required sign. This method is not on the formulae sheet: learn it.

Change only $<$ to $>$. The roots stay fixed, but the chosen region moves outside them.

With $\leqslant0$, include the roots in the inside interval. With $\geqslant0$, include them in the outside intervals. The root positions and signs between them do not change.

Try new roots and an inclusive sign.

Rearranging a quadratic inequality (Higher)

Before choosing intervals, put every term on one side so that the other side is zero. You can then read the sign of one expression. Moving terms by addition or subtraction preserves the direction; multiplying the entire inequality by a negative number reverses it.

For $x^2<3x+4$, compare $x^2-3x-4$ with zero. For $-x^2+3x+4>0$, it is often easier to multiply by $-1$ and compare $x^2-3x-4$ with zero instead. These are applications of the inequality rules, not extra formulae on the sheet.

Change only the presentation: write the same comparison with the quadratic subtracted on the other side. This creates a negative leading coefficient.

A quadratic need not have two distinct roots. For $(x-2)^2\leqslant0$, a square cannot be negative, so the only solution is $x=2$. For $(x-2)^2<0$, there is no real solution. For $x^2+1>0$, every real $x$ works. Check the graph or the expression rather than forcing an “inside or outside” pattern where there are no two distinct roots.

The following inequality has its terms on both sides. Bring them together before using the roots.

Writing solution sets (Higher)

Set notation describes the same allowed values without a drawing. Read $\{x:x>3\}$ as “the set of values $x$ such that $x>3$”. The colon means “such that”. Unless a question states otherwise here, $x$ is real, so fractions and decimals are allowed too.

A double inequality describes an intersection: both limits hold. Two outward intervals describe a union: either interval is enough. State any restriction such as “$x$ is an integer” before selecting values. This notation is not on the formulae sheet: learn it.

Change only the operation from intersection to union. Now either condition can admit a value.

For two disjoint ranges, union does not fill the gap. For example, $\{x:x<-2\}\cup\{x:x>4\}=\{x:x<-2\text{ or }x>4\}$. Its intersection is $\varnothing$, because no real number satisfies both. At the other extreme, $(x-2)^2\leqslant0$ has the solution set $\{2\}$: one value, not an interval.