The quadratic equation in the image is:
$$y^2 – 6y – 8 = 0$$
Step 1: Check for simple integer factors
To factorise a quadratic of the form $y^2 + by + c = 0$ into $(y + p)(y + q) = 0$:
- Two numbers $p$ and $q$ must multiply to give $c = -8$
- The same numbers must add up to give $b = -6$
The integer factor pairs of $-8$ are:
- $1 \times (-8) \longrightarrow 1 + (-8) = -7$
- $(-1) \times 8 \longrightarrow -1 + 8 = 7$
- $2 \times (-4) \longrightarrow 2 + (-4) = -2$
- $(-2) \times 4 \longrightarrow -2 + 4 = 2$
None of these pairs add up to $-6$. Therefore, this expression cannot be factorised using whole numbers (integers).
Step 2: Factorise by Completing the Square
To factorise over real numbers, rewrite the quadratic into vertex form first:
- Take half of the coefficient of $y$ and square it:
$$\text{Half of } -6 = -3 \implies (-3)^2 = 9$$ - Add and subtract $9$ to rewrite the expression:
$$y^2 – 6y + 9 – 9 – 8 = 0$$ - Group the perfect square trinomial:
$$(y – 3)^2 – 17 = 0$$ - Apply the Difference of Two Squares identity ($A^2 – B^2 = (A – B)(A + B)$):
$$(y – 3)^2 – (\sqrt{17})^2 = 0$$$$\bigl(y – 3 – \sqrt{17}\bigr)\bigl(y – 3 + \sqrt{17}\bigr) = 0$$
Step 3: Solve for $y$
| Multivariable Function / System | Yielded Quadratic Equation(s) | Source Context |
| $f(x, y) = x^3 + 3x^2 – 3xy^2 + 18xy – 24x + 1$ | $y^2 – 6y + 8 = 0$ $x^2 + 2x + 1 = 0$ | MST224 2014 Exam, Q7 |
| $f(x, y) = 2x^2 + xy^2 – 6xy + 5x$ | $y^2 – 6y + 5 = 0$ | MST224 2015 Exam, Q21 |
| $f(x, y) = 2x^2 + xy^2 + 6xy + 5x + 2$ | $y^2 + 6y + 5 = 0$ | MST224 2016 Exam, Q15 |
| $f(x, y) = x^2 + xy^2 – 6xy + 5x$ | $y^2 – 6y + 5 = 0$ | MST224 2018 Exam, Q21 |
| $f(x, y) = 2x^3 – 9x^2 – 3xy^2 + 12xy$ | $y^2 – 4y = 0$ $x^2 – 3x + 2 = 0$ | MST224 2019 Exam, Q7 |
| $\begin{cases} f_x = 4x + y^2 – 6y + 5 = 0 \\ f_y = 2xy – 6x = 0 \end{cases}$ | $y^2 – 6y + 5 = 0$ | MST224 2022 Exam, Q30 |
| $\begin{cases} f_x = 3x^2 – 12x – 3y^2 + 12 = 0 \\ f_y = -6xy + 6y = 0 \end{cases}$ | $x^2 – 4x + 4 = 0$ $y^2 – 1 = 0$ | MST224 2023 Exam, Q12 |
| $\begin{cases} f_x = 36x + y^2 – 4y – 32 = 0 \\ f_y = 2xy – 4x = 0 \end{cases}$ | $y^2 – 4y – 32 = 0$ | MST224 2023 Exam, Q30 |
| $\begin{cases} \dot{x} = -4y + 2xy – 8 = 0 \\ \dot{y} = 4y^2 – x^2 = 0 \end{cases}$ | $y^2 – y – 2 = 0$ $y^2 + y + 2 = 0$ | Book 4 (Unit 13), Example 6 |
| $\begin{cases} \dot{x} = (1 + x – 2y)x = 0 \\ \dot{y} = (x – 1)y = 0 \end{cases}$ | $x^2 + x = 0$ | Book 4 (Unit 13), Exercise 25 |
| $\begin{cases} \dot{x} = x(20 – y) = 0 \\ \dot{y} = y(10 – y)(10 – x) = 0 \end{cases}$ | $y^2 – 10y = 0$ | Book 4 (Unit 13), Exercise 9 |
| $\begin{cases} \dot{x} = 0.5x – 0.00005x^2 = 0 \\ \dot{y} = -0.1y + 0.0004xy – 0.01y^2 = 0 \end{cases}$ | $0.01y^2 + 0.1y = 0$ $0.01y^2 – 3.9y = 0$ | Book 4 (Unit 13), Exercise 14 |
Set each factor equal to zero:
$$y – 3 – \sqrt{17} = 0 \implies y = 3 + \sqrt{17}$$
$$y – 3 + \sqrt{17} = 0 \implies y = 3 – \sqrt{17}$$
Final Solutions:
$$y = 3 \pm \sqrt{17}$$