Chapter 12: Review

The chapter problem, solved

The ball reaches the ground when its height is zero:

Factor out $-1$, then factorise the quadratic:

$$ -(t-5)(t+1)=0 $$ So $t=5$ or $t=-1$. Time cannot be negative in this context, so the ball reaches the ground after 5 seconds. Substitution gives $h=0$ when $t=5$.

Mixed practice — choose the method first

For each question, name a method before working it out. Give a reason for your choice.

1. Expand $(x+4)(x-2)$.
2. Factorise $x^2+3x-10$.
3. Solve $x^2-9=0$ by factorising.
4. Solve $3x^2+x-2=0$ using the quadratic formula.
5. Write $x^2+10x+1$ in completed-square form. State its turning point as a graph of $y$.
6. A quadratic graph has roots $-1$ and $4$. What equation could describe its roots?
7. From earlier work: solve $3x+8=20$. Explain how you checked your answer.
8. From earlier work: simplify $2^3\times2^4$.

What you can now do

• expand and factorise quadratic expressions;
• read roots and turning points from equations and graphs;
• solve quadratic equations with three methods and choose a suitable one.

Your relevance reflection

Return to your example from the chapter opener. Explain how a quadratic model could help answer a question in that situation. If it does not fit, say what kind of quantity would need to change quadratically.