IB Mathematics AA HL Flashcards- Roots of Quadratics

FACTORISATION ➔ We saw in 2.6 that it was easy to find the roots of a quadratic if it was in the form: \(f(x) = a(x – p)(x – q)\), because you just need to make each bracket equal zero. But as you will often be given \(f(x) = ax^2 + bx + c\) initially, you need a process for changing forms – factorisation.

[q]

WHEN a = 1 ➔ If you take \(f(x) = (x – 2)(x + 3)\), we can expand:
\[
f(x) = x^2 – 2x + 3x – 6 = x^2 + 1x – 6
\]

[a]

We notice that the \(-6\) was found by multiplying \(-2\) & 3, and the \(1\) \([from \, 1x]\) was found by adding: \(-2 + 3 = 1\).
Therefore, if we want to move in the other direction, we must find two numbers that multiply to get c, and sum to get b.

[q]

E.G.  ➔ Factorise \(x^2 + 9x + 18\):
\(9 \times 2 = 18\), but \(9 + 2 = 9\).
\(6 \times 3 = 18\), \(6 + 3 = 9\) ✔:
\[
(x + 6)(x + 3)
\]

[a]

WHEN a ≠ 1 ➔ This makes the process much harder to pick the \(p\) & \(q\). Firstly, you can check if the whole equation is divisible by ‘a’, then you can just do the above method after factoring the whole thing by ‘a’. But sometimes, we must go from \(ax^2 + bx + c \rightarrow (mx + p)(nx + q)\). Here \(p \cdot q\) still equals ‘c’, but ‘b’ would equal \((mq + np)\), and ‘a’ would equal \(mn\). This is sometimes possible through observation/testing:

[q]

E.G.  ➔ Factorise \(3x^2 + 11x + 10\):
Firstly \(m\) & \(n\) must be 3 & 1 (or \(-3\) & \(-1\)). So now, we need \(3q + 1p = 11\) w/ \(pq = 10\).
Try \(10 \& 1\):
\(3(10) + 1 \neq 11\), and \(3(1) + 10 \neq 11\).
Try \(5 \& 2\):
\(3(5) + 2 = 11\), but \(3(2) + 5 = 11\):
\[
3x^2 + 11x + 10 = (3x + 5)(x + 2)
\]

[a]

There are a few quicker methods. Here’s one that is very compact/efficient:

FUNNEL METHOD (from E.G.3):
\[
\begin{aligned}
&3 \, \vert \, 11 \, \vert \, 10 \quad &\text{(take coefficients)} \\
&1 \quad 2 \quad &\text{(2 numbers with prod. 30 \& sum 11)} \\
&6 \quad 5 \quad &\text{(fractions over a)} \\
\end{aligned}
\quad \text{simplify}
\]

\[
= (2x + 2)(3x + 5)
\]

[q]

SOLVING/ROOTS ➔ So, as we know, after factoring, solve either bracket equal to zero to find 2 roots to the quadratic equation. There are other ways:

COMPLETING THE SQUARE ➔ This is the process of going from \(ax^2 + bx + c \rightarrow a(x – h)^2 + k\).

[a]

If you expand \(a(x – h)^2 + k\), you get \(ax^2 – 2ha x + (ah^2 + k)\), so \(b = -2ha \Rightarrow h = \frac{-b}{2a}\) [like the Ax. of Symm.], and \(k\) is a correction from \(ah^2\) to get \(c\). A benefit of \(a(x – h)^2 + k\) is that it’s rearrangeable, so you can solve to find the roots:

[q]
E.G.  ➔ Solve \(4x^2 – 4x – 15 = 0\) by completing the square (same Q as E.G.):
\(h = \frac{-b}{2a} = \frac{-(-4)}{2(4)} = \frac{1}{2}\), \(ah^2 = 1\) and should be \(-15\), so \(k = -16\):
\(4(x – \frac{1}{2})^2 – 16 = 0 \Rightarrow 4(x – \frac{1}{2})^2 = 16 \Rightarrow (x – \frac{1}{2})^2 = 4 \Rightarrow x – \frac{1}{2} = 2 \quad x = 2.5 \& 1.5\)

[a]

QUADRATIC FORMULA ➔ We can now actually generalise what we did in the final line of E.G.6&7. We had \(a(x – h)^2 + k = a(x + \frac{b}{2a})^2 + c\), and
\(a(\frac{x + b}{2a})^2 – c = \frac{b^2}{a^2} – c \Rightarrow \frac{x + b}{2a} = \pm \sqrt{\frac{b^2}{a^2} – c}\)
This is a formula to find roots!
\[
x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}
\]
in F.B.

[q]

E.G.  ➔ Solve \(2x^2 – 3x – 5 = 0\) using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} = \frac{3 \pm \sqrt{9 + 40}}{4} = \frac{3 \pm \sqrt{49}}{4} = \frac{3 \pm 7}{4} \Rightarrow \frac{3 + 7}{4} \, \text{or} \, \frac{3 – 7}{4} \Rightarrow 2.5 \, \text{or} \, -1
\]

[a]

NO. OF ROOTS ➔ This formula is also helpful for identifying when we don’t get ‘two roots’. This is because you can’t do the square root of a negative, and \(b^2 – 4ac\) can be negative. If 0 are 2 of the same answers, giving us a repeated root. This part of the quadratic formula is called the discriminant (\(\Delta\)) i.e.:
\[
\Delta = b^2 – 4ac
\]
in F.B.

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