Quadratic inequalities

A quadratic inequality asks where a function is positive or negative. $\displaystyle x^{2}-4>0$ asks: for which $\displaystyle x$ is the function $\displaystyle {f(x)}={x^{2}-4}$ positive – where is its graph above the $\displaystyle x$-axis? We already know how to read the positive and negative domains from a graph; here we find them by calculation and a sketch.

The graph of $\displaystyle {f(x)}={x^{2}-4}$ meets the $\displaystyle x$-axis at $\displaystyle -2$ and at $\displaystyle 2$ – its roots, the $\displaystyle x$ values where $\displaystyle {f(x)}={0}$.

• $\displaystyle x^{2}-4<0$ – the graph is below the axis: $\displaystyle -2<x<2$.
• $\displaystyle x^{2}-4>0$ – the graph is above the axis: $\displaystyle x<-2$ or $\displaystyle x>2$.
• $\displaystyle x^{2}-4\geq0$ – above the axis or on it: $\displaystyle x\leq-2$ or $\displaystyle x\geq2$ (the roots are included).

1. Bring everything to one side, so that the other side is 0.
2. Find the roots: solve the equation $\displaystyle {f(x)}={0}$.
3. Sketch the parabola through the roots: $\displaystyle a>0$ – it opens upwards, $\displaystyle a<0$ – downwards.
4. Read the answer: $\displaystyle >0$ – where the graph is above the axis, $\displaystyle <0$ – where it is below it.

$\displaystyle x^{2}-5x+6\geq0$.
The roots: $\displaystyle {(x-2)(x-3)}={0}$, so $\displaystyle {x}={2}$ or $\displaystyle {x}={3}$. Here $\displaystyle {a}={1>0}$, so the parabola opens upwards: it is below the axis between the roots and above it outside them. The answer: $\displaystyle x\leq2$ or $\displaystyle x\geq3$.

$\displaystyle -x^{2}+2x+3>0$.
The roots: $\displaystyle {-x^{2}+2x+3}={0}$, multiplied by $\displaystyle -1$: $\displaystyle {x^{2}-2x-3}={0}$, $\displaystyle {(x-3)(x+1)}={0}$, so $\displaystyle {x}={3}$ or $\displaystyle {x}={-1}$. Here $\displaystyle {a}={-1<0}$, so the parabola opens downwards: it is above the axis between the roots. The answer: $\displaystyle -1<x<3$.

From $\displaystyle x^{2}<4$ it is easy to write “$\displaystyle x<2$”. But $\displaystyle {x}={-5}$ satisfies $\displaystyle x<2$, and $\displaystyle {(-5)^{2}}={25}$ is not less than 4. The right answer comes from the graph: $\displaystyle -2<x<2$.

Here is the graph of $\displaystyle {f(x)}={x^{2}+2x-3}$.
Find its roots by factoring, and mark them on the graph.
Write for which $\displaystyle x$:
a. $\displaystyle f(x)<0$
b. $\displaystyle f(x)\geq0$

Solve the inequalities: find the roots, sketch the parabola, and read the answer.

When both sides have terms, first bring everything to one side: $\displaystyle x^{2}>x+2$ becomes $\displaystyle x^{2}-x-2>0$, that is $\displaystyle (x-2)(x+1)>0$. The parabola opens upwards, so the answer is outside the roots: $\displaystyle x<-1$ or $\displaystyle x>2$.

Solve.
Bring everything to one side first.

A parabola that does not cross the $\displaystyle x$-axis is all on one side of it.

• $\displaystyle {f(x)}={x^{2}+2}$ has no roots and opens upwards, so it is always above the axis: $\displaystyle x^{2}+2>0$ holds for every $\displaystyle x$, and $\displaystyle x^{2}+2<0$ has no solution.
• $\displaystyle {f(x)}={x^{2}-4x+4}={(x-2)^{2}}$ touches the axis only at its vertex, $\displaystyle {x}={2}$: $\displaystyle (x-2)^{2}>0$ holds for every $\displaystyle x$ except 2.

Solve, and explain with a sketch.
a. $\displaystyle x^{2}+5>0$
b. $\displaystyle x^{2}+5<0$
c. $\displaystyle (x-1)^{2}\leq0$
d. $\displaystyle -x^{2}-1<0$