Standard +0.8 This question requires understanding discriminants for two linked quadratic equations, deriving an inequality from the first condition (b² - 4ac > 0 gives m < 9), then proving the second equation's discriminant is negative. It involves multi-step algebraic manipulation and logical reasoning connecting two equations, going beyond routine discriminant calculations but remaining within standard A-level techniques.
The quadratic equation \(4x^2 - 12x + m = 0\), where \(m\) is a positive constant, has two distinct real roots.
Show that the quadratic equation \(3x^2 + mx + 7 = 0\) has no real roots. [7]
An expression for \(b^2 – 4ac\) for the quadratic equation \(4x^2 – 12x + m = 0\), with at least two of \(a\), \(b\) or \(c\) correct
M1
AO1
\(b^2 – 4ac = 12^2 – 4 \times 4 \times m\)
A1
AO1
\(b^2 – 4ac > 0\)
m1
AO1
\((0<) m < 9\)
A1
AO1
An expression for \(b^2 – 4ac\) for the quadratic equation \(3x^2 + mx + 7 = 0\), with at least two of \(a\), \(b\) or \(c\) correct
(M1)
— (to be awarded only if the corresponding M1 is not awarded above)
\(b^2 – 4ac = m^2 – 84\)
A1
AO2
\(m^2 < 81 \Rightarrow b^2 – 4ac < -3\)
A1
AO2
\(b^2 – 4ac < 0 \Rightarrow\) no real roots
A1
AO2
Total: [7]
An expression for $b^2 – 4ac$ for the quadratic equation $4x^2 – 12x + m = 0$, with at least two of $a$, $b$ or $c$ correct | M1 | AO1
$b^2 – 4ac = 12^2 – 4 \times 4 \times m$ | A1 | AO1
$b^2 – 4ac > 0$ | m1 | AO1
$(0<) m < 9$ | A1 | AO1
An expression for $b^2 – 4ac$ for the quadratic equation $3x^2 + mx + 7 = 0$, with at least two of $a$, $b$ or $c$ correct | (M1) | — (to be awarded only if the corresponding M1 is not awarded above)
$b^2 – 4ac = m^2 – 84$ | A1 | AO2
$m^2 < 81 \Rightarrow b^2 – 4ac < -3$ | A1 | AO2
$b^2 – 4ac < 0 \Rightarrow$ no real roots | A1 | AO2
**Total: [7]**
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The quadratic equation $4x^2 - 12x + m = 0$, where $m$ is a positive constant, has two distinct real roots.
Show that the quadratic equation $3x^2 + mx + 7 = 0$ has no real roots. [7]
\hfill \mbox{\textit{WJEC Unit 1 Q9 [7]}}