In the quadratic equation \(7{x^2} - 8x + p = 0,{\text{ }}(p \in \mathbb{Q})\), one root is three times the other root.
Find the value of \(p\).

METHOD 1

let roots be \(\alpha \) and \(3\alpha \)     (M1)

sum of roots \((4\alpha ) = \frac{8}{7}\)     M1

\( \Rightarrow \alpha  = \frac{2}{7}\)     A1

EITHER

product of roots \((3{\alpha ^2}) = \frac{p}{7}\)     M1

\(p = 21{\alpha ^2} = 21 \times \frac{4}{{49}}\)

OR

\(7{\left( {\frac{2}{7}} \right)^2} - 8\left( {\frac{2}{7}} \right) + p = 0\)     M1

\(\frac{4}{7} - \frac{{16}}{7} + p = 0\)

THEN

\( \Rightarrow p = \frac{{12}}{7}{\text{ }}( = 1.71)\)     A1

METHOD 2

\(x = \frac{{8 \pm \sqrt {64 - 28p} }}{{14}}\)     (M1)

\(\frac{{8 + \sqrt {64 - 28p} }}{{14}} = 3\left( {\frac{{8 - \sqrt {64 - 28p} }}{{14}}} \right)\)     M1A1

\(8 + \sqrt {64 - 28p}  = 24 - 3\sqrt {64 - 28p}  \Rightarrow \sqrt {64 - 28p}  = 4\)     (M1)

\(p = \frac{{12}}{7}{\text{ }}( = 1.71)\)     A1

[5 marks]