Moderate -0.8 This is a straightforward discriminant question requiring students to apply b²-4ac < 0 for no real roots, then solve a simple quadratic inequality. It's a standard AS-level technique with minimal problem-solving required, making it easier than average but not trivial due to the inequality manipulation needed.
4 The equation \(9 x ^ { 2 } + 4 x + p ^ { 2 } = 0\) has no real solutions for \(x\).
Find the set of possible values of \(p\).
Fully justify your answer. [0pt]
[4 marks]
For no real solutions the discriminant must be negative
B1
AO 1.2
\(4^2 - 4 \times 9 \times p^2 < 0\) leading to \(p^2 > \frac{4}{9}\)
M1
AO 1.1a — substitutes \(b=4\), \(a=9\), \(c=p^2\) into \(b^2 - 4ac\)
Critical values \(\frac{2}{3}\) and \(-\frac{2}{3}\)
A1
AO 2.2a
\(p > \frac{2}{3}\) or \(p < -\frac{2}{3}\)
A1
AO 2.5 — two correct inequalities for \(p\)
## Question 4:
| Answer/Working | Mark | Guidance |
|---|---|---|
| For no real solutions the discriminant must be negative | B1 | AO 1.2 |
| $4^2 - 4 \times 9 \times p^2 < 0$ leading to $p^2 > \frac{4}{9}$ | M1 | AO 1.1a — substitutes $b=4$, $a=9$, $c=p^2$ into $b^2 - 4ac$ |
| Critical values $\frac{2}{3}$ and $-\frac{2}{3}$ | A1 | AO 2.2a |
| $p > \frac{2}{3}$ or $p < -\frac{2}{3}$ | A1 | AO 2.5 — two correct inequalities for $p$ |
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4 The equation $9 x ^ { 2 } + 4 x + p ^ { 2 } = 0$ has no real solutions for $x$.
Find the set of possible values of $p$.\\
Fully justify your answer.\\[0pt]
[4 marks]\\
\hfill \mbox{\textit{AQA AS Paper 2 2022 Q4 [4]}}