This checkpoint assesses your understanding of quadratic equations and roots, the null factor law, solving $x^2 + bx + c = 0$ by factoring (trinomials and the difference of two squares), and the link between roots and $x$-intercepts. It covers Lessons 18–20.
❓ Multiple Choice (10 questions)
1. Solve $x^2 = 49$.
2. The solutions of $(x - 4)(x + 1) = 0$ are:
3. According to the null factor law, if $A \times B = 0$ then:
4. Factorise $x^2 - 5x + 6$.
5. Solve $x^2 - 9 = 0$.
6. Solve $x^2 + x - 6 = 0$.
7. The $x$-intercepts of the parabola $y = x^2 - 7x + 12$ are at:
8. How many real solutions does $x^2 + 4 = 0$ have?
9. Solve $x^2 + 2x - 15 = 0$.
10. A parabola $y = x^2 + bx + c$ cuts the $x$-axis at $x = -2$ and $x = 5$. Which equation has these roots?
✍ Short Answer (4 questions)
11. Solve by inspection (remember $\pm$).
(a) $x^2 = 36$ (1 mark)
(b) $x^2 - 100 = 0$ (2 marks)3 MARKS
12. Solve each quadratic equation by factoring.
(a) $(x - 3)(x + 5) = 0$ (1 mark)
(b) $x^2 - 8x + 15 = 0$ (2 marks)
(c) $x^2 + 3x - 10 = 0$ (2 marks)5 MARKS
13. Roots and $x$-intercepts.
(a) Find the $x$-intercepts of $y = x^2 - 6x + 8$. (2 marks)
(b) Verify your answers by substituting one of them back into the equation. (1 mark)3 MARKS
14. Synthesis.
(a) Solve $x^2 - 4x = 0$ by factoring (take out the common factor). (2 marks)
(b) A ball's height is modelled by $h = -t^2 + 6t$, where $h$ is in metres and $t$ in seconds. Find the times when the ball is at ground level ($h = 0$), and state how long the ball is in the air. (3 marks)5 MARKS
1. C$x^2 = 49 \Rightarrow x = \pm 7$, so $x = 7$ or $x = -7$. Both square to 49.
2. BBy the null factor law, $x - 4 = 0$ or $x + 1 = 0$, giving $x = 4$ or $x = -1$.
3. AThe null factor law: if a product is zero, at least one factor is zero, so $A = 0$ or $B = 0$.
4. CTwo numbers multiplying to $+6$ and adding to $-5$ are $-2$ and $-3$: $(x - 2)(x - 3)$.
5. B$x^2 - 9 = (x - 3)(x + 3) = 0$ (difference of two squares), so $x = 3$ or $x = -3$.
6. A$x^2 + x - 6 = (x + 3)(x - 2) = 0$, so $x = -3$ or $x = 2$.
7. D$x^2 - 7x + 12 = (x - 3)(x - 4) = 0$, so the $x$-intercepts are at $x = 3$ and $x = 4$.
8. B$x^2 = -4$ has no real solution (a square can't be negative), so there are none.
9. C$x^2 + 2x - 15 = (x + 5)(x - 3) = 0$, so $x = -5$ or $x = 3$.
10. ARoots $-2$ and $5$ come from factors $(x + 2)$ and $(x - 5)$: $(x + 2)(x - 5) = 0$.
Q11 (3 marks): (a) $x^2 = 36 \Rightarrow x = \pm 6$ [1]. (b) $x^2 = 100 \Rightarrow x = \pm 10$ (i.e. $x = 10$ or $x = -10$) [2].
Q12 (5 marks): (a) $x - 3 = 0$ or $x + 5 = 0 \Rightarrow x = 3$ or $x = -5$ [1]. (b) $(x - 3)(x - 5) = 0 \Rightarrow x = 3$ or $x = 5$ [2]. (c) $(x + 5)(x - 2) = 0 \Rightarrow x = -5$ or $x = 2$ [2].
Q13 (3 marks): (a) Set $y = 0$: $x^2 - 6x + 8 = (x - 2)(x - 4) = 0 \Rightarrow x = 2$ or $x = 4$, so $x$-intercepts $(2, 0)$ and $(4, 0)$ [2]. (b) Check $x = 2$: $(2)^2 - 6(2) + 8 = 4 - 12 + 8 = 0$ ✓ [1].
Q14 (5 marks): (a) $x^2 - 4x = x(x - 4) = 0 \Rightarrow x = 0$ or $x = 4$ [2]. (b) $-t^2 + 6t = 0 \Rightarrow -t(t - 6) = 0 \Rightarrow t = 0$ or $t = 6$. The ball is at ground level at $t = 0$ s (launch) and $t = 6$ s (landing), so it is in the air for 6 seconds [3].
Tick when you have finished all questions and checked your answers.