2.6 Linear and quadratic inequalities, and set notation — Practise with help

1. $4x = 12$, so $x = 3$.
2. $x^2 - 5x + 4 = (x - 1)(x - 4)$.
3. From the factorisation, $x = 1$ or $x = 4$.

If question 3 was hard, revisit factorising: the roots of the equation are the same numbers that appear in the brackets.

Subtract $2$ from both sides, then divide by $5$. Both operations keep the direction of the inequality.

After subtracting, you have $5x > 15$. Now divide by the positive number $5$.

Subtract $2$:

Divide by $5$:

The solution set is $x \in (3, \infty)$. Check with $x = 4$: $5 \times 4 + 2 = 22 > 17$.

Dividing by $-3$ reverses the direction of the inequality. Use a test value to check.

Divide both sides by $-3$ and turn $\leq$ into $\geq$.

Divide by $-3$, reversing the sign:

The solution set is $x \in [-4, \infty)$. Check with $x = 0$: $-3 \times 0 = 0 \leq 12$, which is true, and $0 \geq -4$ is also true.

Find the roots first, then sketch the parabola. The coefficient of $x^2$ is positive, so it opens upwards.

Factorise $x^2 - 4x - 5 = (x - 5)(x + 1)$ to get the roots $x = -1$ and $x = 5$.

Factorise:

so the parabola crosses the $x$-axis at $x = -1$ and $x = 5$. It opens upwards, so it is below the axis between the roots. The inequality asks for $< 0$:

The solution set is $x \in (-1, 5)$.

Rearrange to $x^2 - 16 \geq 0$ and factorise a difference of two squares. The answer has two pieces.

$(x - 4)(x + 4) \geq 0$. The parabola is positive outside the roots, so write the answer as a union.

Rearrange and factorise:

The parabola opens upwards and the inequality includes zero, so the solution is outside the roots:

In set notation, $x \in (-\infty, -4] \cup [4, \infty)$.

The coefficient of $x^2$ is $2$, so factorise carefully. Look for a factor pair that gives a middle term of $3x$.

$2x^2 + 3x - 2 = (2x - 1)(x + 2)$, so the roots are $x = \frac{1}{2}$ and $x = -2$.

Factorise:

The roots are $x = -2$ and $x = \frac{1}{2}$, and the parabola opens upwards. The inequality asks for $\leq 0$, which is between the roots:

The solution set is $x \in \left[-2, \frac{1}{2}\right]$.

"At least $1$ m" means $h \geq 1$, so set $1 + 4t - 4t^2 \geq 1$ and simplify.

Subtract $1$ from both sides: $4t - 4t^2 \geq 0$. Factorise the left side.

Set $h \geq 1$:

Factorise:

The expression is a quadratic with negative $t^2$ coefficient, so it opens downwards. It is positive between the roots $t = 0$ and $t = 1$. The ball is at least $1$ m high for

The model is meant for $t \geq 0$, so the realistic answer is $t \in [0, 1]$.