21N.1.AHL.TZ0.7

pestleMathematicsAAHLPaper 121N· sl-2-7-solutions-of-quadratic-equations-and-inequalities-discriminant-and-nature-of-rootssource ↗

The equation $3 p x^{2} + 2 p x + 1 = p$ has two real, distinct roots.

Find the possible values for $p$ .

[5]

Consider the case when $p = 4$ . The roots of the equation can be expressed in the form $x = \frac{a \pm \sqrt{13}}{6}$ , where $a \in ℤ$ . Find the value of $a$ .

[2]

Markscheme / solution

attempt to use discriminant $b^{2} - 4 a c (> 0)$ M1

${(2 p)}^{2} - 4 (3 p) (1 - p) (> 0)$

$16 p^{2} - 12 p (> 0)$ (A1)

$p (4 p - 3) (> 0)$

attempt to find critical values $(p = 0, p = \frac{3}{4})$ M1

recognition that discriminant $> 0$ (M1)

$p < 0$ or $p > \frac{3}{4}$ A1

Note: Condone ‘or’ replaced with ‘and’, a comma, or no separator

[5 marks]

$p = 4 \Rightarrow 12 x^{2} + 8 x - 3 = 0$

valid attempt to use $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ (or equivalent) M1

$x = \frac{- 8 \pm \sqrt{208}}{24}$

$x = \frac{- 2 \pm \sqrt{13}}{6}$

$a = - 2$ A1

[2 marks]

Examiners’ report

[N/A]