Standard +0.3 This is a straightforward application of the quadratic formula with complex coefficients. While it requires careful arithmetic with complex numbers and simplification of a complex square root, it follows a standard procedure without requiring novel insight or extended problem-solving. Slightly above average difficulty due to the algebraic manipulation involved, but well within the scope of routine P3/Further Maths questions.
4 Solve the quadratic equation \(( 3 + \mathrm { i } ) w ^ { 2 } - 2 w + 3 - \mathrm { i } = 0\), giving your answers in the form \(x + \mathrm { i } y\), where \(x\) and \(y\) are real.
Obtain one of the answers \(w = \dfrac{2+6i}{6+2i}\) or \(w = \dfrac{2-6i}{6+2i}\)
A1
Must be simplified to this form.
Show intention to multiply numerator and denominator by conjugate of their denominator.
M1
Independent of previous M marks but must be of the same form, e.g. \(\dfrac{a}{b+ci}\).
Obtain final answers \(\dfrac{3}{5} + \dfrac{4}{5}i\) and \(-i\). Accept \(0.6 + 0.8i\) and \(0 - i\)
A1
SC Both correct final answers from \(w = \dfrac{2+6i}{6+2i}\) and \(w = \dfrac{2-6i}{6+2i}\) seen, no evidence of conjugate, then SC B1 for both. Allow \(x = \frac{3}{5}\), \(y = \frac{4}{5}\) or \(x = 0\), \(y = -1\). A0 for \(\dfrac{3+4i}{5}\).
Alternative: Multiply the equation by \(3-i\)
M1
Use \(i^2 = -1\) in \((3+i)(3-i)\)
M1
Obtain \(10w^2 - 2(3-i)w + (3-i)^2 = 0\) or equivalent
A1
Use quadratic formula or factorise to solve for \(w\)
M1
Obtain final answers \(\dfrac{3}{5} + \dfrac{4}{5}i\) and \(-i\)
A1
SC Both correct final answers from \(10w^2 - 2(3-i)w + (3-i)^2 = 0\) with no working then SC B1 for both.
Alternative: Substitute \(w = x + iy\) and form equations for real and imaginary parts
Form quartic equation in \(x\) only or \(y\) only using the correct substitution and solve for \(x\) or \(y\)
M1
Obtain final answers \(\dfrac{3}{5} + \dfrac{4}{5}i\) and \(-i\)
A1
SC Both correct final answers with no working then SC B1 for both.
5
## Question 4:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{2 \pm \sqrt{(-2)^2 - 4(3+i)(3-i)}}{2(3+i)}$ | M1 | Use quadratic formula to solve for $w$ |
| Use $i^2 = -1$ in $(3+i)(3-i)$ | M1 | |
| Obtain one of the answers $w = \dfrac{2+6i}{6+2i}$ or $w = \dfrac{2-6i}{6+2i}$ | A1 | Must be simplified to this form. |
| Show intention to multiply numerator and denominator by conjugate of their denominator. | M1 | Independent of previous M marks but must be of the same form, e.g. $\dfrac{a}{b+ci}$. |
| Obtain final answers $\dfrac{3}{5} + \dfrac{4}{5}i$ and $-i$. Accept $0.6 + 0.8i$ and $0 - i$ | A1 | SC Both correct final answers from $w = \dfrac{2+6i}{6+2i}$ and $w = \dfrac{2-6i}{6+2i}$ seen, no evidence of conjugate, then SC B1 for both. Allow $x = \frac{3}{5}$, $y = \frac{4}{5}$ or $x = 0$, $y = -1$. A0 for $\dfrac{3+4i}{5}$. |
| **Alternative:** Multiply the equation by $3-i$ | M1 | |
| Use $i^2 = -1$ in $(3+i)(3-i)$ | M1 | |
| Obtain $10w^2 - 2(3-i)w + (3-i)^2 = 0$ or equivalent | A1 | |
| Use quadratic formula or factorise to solve for $w$ | M1 | |
| Obtain final answers $\dfrac{3}{5} + \dfrac{4}{5}i$ and $-i$ | A1 | SC Both correct final answers from $10w^2 - 2(3-i)w + (3-i)^2 = 0$ with no working then SC B1 for both. |
| **Alternative:** Substitute $w = x + iy$ and form equations for real and imaginary parts | M1 | |
| Use $i^2 = -1$ in $(x+iy)^2$ | M1 | |
| Obtain $3(x^2 - y^2) - 2xy - 2x + 3 = 0$ and $x^2 - y^2 + 6xy - 2y - 1 = 0$ | A1 | OE |
| Form quartic equation in $x$ only or $y$ only using the correct substitution and solve for $x$ or $y$ | M1 | |
| Obtain final answers $\dfrac{3}{5} + \dfrac{4}{5}i$ and $-i$ | A1 | SC Both correct final answers with no working then SC B1 for both. |
| | **5** | |
4 Solve the quadratic equation $( 3 + \mathrm { i } ) w ^ { 2 } - 2 w + 3 - \mathrm { i } = 0$, giving your answers in the form $x + \mathrm { i } y$, where $x$ and $y$ are real.\\
\hfill \mbox{\textit{CAIE P3 2023 Q4 [5]}}