Quadratic and Rational Inequalities

Step 1Step 1, make $a$ positive: is the leading coefficient positive?

$a=1>0$: the parabola opens up.

✗ Multiply both sides by $-1$:$a=1$ is already positive; multiplying by $-1$ would only force you to flip the sign.

Step 2Step 2, the roots of the matching equation:

$(x+2)(x-3)=0$: roots $-2,\ 3$.

✗ $x=2,\ x=-3$:$x^2-x-6=(x+2)(x-3)$, so the roots are $-2$ and $3$ — mind the signs: $x+2=0$ gives $x=-2$.

✗ No real root:$\Delta=1+24=25>0$: two roots.

Step 3Step 3, read the graph: which part?

Greater than: outside.

✗ between the roots:An upward parabola is above the axis outside the roots — "greater than: outside".

Step 4Step 4, the endpoints:

Strict: open intervals.

✗ Include them (closed):The inequality is strict; at a root the value is $0$, which is not $>0$.

Step 1Step 1, make $a$ positive: is the leading coefficient positive?

It becomes $x^2-2x-3\leq 0$.

✗ Multiply by $-1$: $x^2-2x-3\geq 0$:Multiplying by a negative number flips the sign: $\geq$ becomes $\leq$. This is the most common mistake on these questions.

✗ Leave it:$a=-1<0$: it opens down. Make $a$ positive first so one rule covers every case.

Step 2Step 2, the roots of the matching equation:

$(x+1)(x-3)\leq 0$.

✗ $x=1,\ x=-3$:$x^2-2x-3=(x+1)(x-3)$: roots $-1$ and $3$.

Step 3Step 3, read the graph: which part?

Less than: inside.

✗ outside the roots:It is now $\leq 0$: less than takes the inside.

Step 4Step 4, the endpoints:

With equality: closed.

✗ Exclude them (open):With $\leq$, a root gives $0$, which satisfies it: include it.

Step 1Step 1, make $a$ positive: is the leading coefficient positive?

Opens up.

✗ Multiply by $-1$:$a=1$ is already positive.

Step 2Step 2, the roots of the matching equation:

$(x-2)^2>0$, double root $x=2$.

✗ Two roots $x=\pm 2$:$x^2-4x+4=(x-2)^2$, $\Delta=0$: a single double root.

✗ No real root:$\Delta=16-16=0$: one double root.

Step 3Step 3, read the graph: which part?

It only touches the axis at $x=2$.

✗ $\mathbb{R}$:At $x=2$, $(x-2)^2=0$, which is not $>0$.

✗ $\varnothing$:The parabola only touches the axis at $x=2$ and is above it everywhere else.

Step 4Step 4, the endpoints:

Solution: $\{x\mid x\neq 2\}$.

✗ Write $(2,+\infty)$:The part $x<2$ satisfies it too.

Step 1Step 1, convert: the fraction inequality is equivalent to

A quotient has the sign of the product; then drop the denominator's root.

✗ $(x-1)(x+2)\leq 0$:At $x=-2$ the denominator is $0$ — with $\leq$ you must remove the denominator's root separately.

✗ $x-1\leq 0$:You cannot multiply by $x+2$ — it may be negative. A quotient and a product have the same sign: use the product.

Step 2Step 2, the roots of the matching equation:

Roots $-2,\ 1$.

✗ $x=-1,\ x=2$:$x-1=0$ gives $1$, $x+2=0$ gives $-2$.

Step 3Step 3, read the graph: which part?

Less than: inside.

✗ outside the roots:$\leq 0$: inside.

Step 4Step 4, the endpoints:

Numerator roots follow the equality sign; denominator roots are always open.

✗ Both closed:$x=-2$ makes the denominator $0$: never included.

✗ Both open:At $x=1$ the fraction is $0$, which satisfies $\leq 0$.

Step 1Step 1, move everything to one side:

$\frac{x+2}{x-1}-2=\frac{4-x}{x-1}\geq 0$.

✗ $x+2\geq 2(x-1)$:Do not multiply by $x-1$ — its sign is unknown. Move the $2$ over: $\frac{x+2}{x-1}-2=\frac{4-x}{x-1}$.

Step 2Step 2, make the coefficient of $x$ positive:

$\frac{x-4}{x-1}\leq 0$.

✗ $\frac{x-4}{x-1}\geq 0$:Multiplying the numerator by $-1$ flips the sign.

Step 3Step 3, read the graph: which part?

Between $1$ and $4$.

✗ Outside $1$ and $4$:$\leq 0$: inside.

Step 4Step 4, the endpoints:

$(1,4]$.

✗ $1$ closed, $4$ open:Reversed: $x=1$ zeroes the denominator (never included); $x=4$ makes the fraction $0$ (included).