Today's lesson
The quadratic formula solves any quadratic equation — even ones that won't factorise. The piece under the square root, the discriminant, also tells us how many solutions there are before we even solve.
Learning intentions
- Know the quadratic formula and when to apply it
- Use the quadratic formula to solve a quadratic equation (exact, surd answers)
- Know what the discriminant $\Delta=b^2-4ac$ is and what it tells you
- Use the discriminant to determine the number of real solutions
Part 1 — The formula & the discriminant (~8 min)
Key idea
- If $ax^2+bx+c=0$ (with $a\ne 0$), then $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$. This is the quadratic formula.
- The discriminant is $\Delta=b^2-4ac$ — the part under the square root.
- $\Delta>0$ → two real solutions. $\Delta=0$ → one real solution. $\Delta<0$ → no real solutions (a square root of a negative number is undefined).
Building understanding — find the discriminant $\Delta=b^2-4ac$ for each.
- $3x^2+2x+1=0$
- $5x^2+3x-2=0$
- $2x^2-x-5=0$
- $-3x^2+4x-5=0$
b) $\Delta=3^2-4(5)(-2)=49$ (two solutions)
c) $\Delta=(-1)^2-4(2)(-5)=41$ (two solutions)
d) $\Delta=4^2-4(-3)(-5)=-44$ (no solutions)
Part 2 — Using the discriminant (Example 1, ~12 min)
Now you try: Find the number of real solutions of $x^2+8x+16=0$, $3x^2-x+2=0$, $x^2+7x-1=0$. Answers: one ($\Delta=0$); none ($\Delta=-23$); two ($\Delta=53$).
📺 Walkthrough: the quadratic formula and the discriminant $\Delta=b^2-4ac$ — how its sign tells you whether there are two, one, or no real solutions.
Part 3 — Solving with the formula (Example 2, ~14 min)
$x=\dfrac{-5\pm\sqrt{5^2-4(1)(3)}}{2(1)}=\dfrac{-5\pm\sqrt{25-12}}{2}=\boxed{\dfrac{-5\pm\sqrt{13}}{2}}$
$x=\dfrac{-(-2)\pm\sqrt{(-2)^2-4(2)(-1)}}{2(2)}=\dfrac{2\pm\sqrt{4+8}}{4}=\dfrac{2\pm 2\sqrt3}{4}=\boxed{\dfrac{1\pm\sqrt3}{2}}$
Now you try: Solve $x^2+3x+1=0$ and $4x^2-2x-3=0$. Answers: $x=\dfrac{-3\pm\sqrt5}{2}$; $x=\dfrac{1\pm\sqrt{13}}{4}$.
Practice 3.1 — solve with the quadratic formula; leave answers exact.
- $x^2+3x-3=0$
- $2x^2+5x+1=0$
- $3x^2-2x-2=0$
- $x^2-4x+1=0$
b) $x=\dfrac{-5\pm\sqrt{17}}{4}$
c) $x=\dfrac{1\pm\sqrt7}{3}$
d) $x=2\pm\sqrt3$
Part 4 — Quick quiz (5 min)
Pick the correct answer for each, then click Mark.
Q1. The discriminant of $ax^2+bx+c=0$ is:
Q2. If $\Delta>0$, the equation has:
Q3. For $x^2+x+1=0$, $\Delta=1-4=-3$, so the equation has:
Q4. In the formula, for $2x^2-2x-1=0$ the values are:
Q5. Using the formula, $x^2+2x-1=0$ gives:
Working program — Cambridge Ex 5J (p469)
After the quiz, open Cambridge Chapter 5 (page 469) and complete the following:
| Set work | Extension |
|---|---|
| Questions 1–3 (2nd column), 5a,c,e,g,i, 6–9 | Questions 10, 11, 12 (½) |
Write the equation as $ax^2+bx+c=0$ first, then read off $a$, $b$ and $c$. Keep answers exact.
Exit ticket — write in your book
- Write down the quadratic formula from memory.
- What does $\Delta=b^2-4ac$ tell you about a quadratic?
- If $\Delta=0$, how many solutions are there?