The equation \({x^2} - 5x - 7 = 0\) has roots \(\alpha \) and \(\beta \). The equation \({x^2} + px + q = 0\) has roots \(\alpha  + 1\) and \(\beta  + 1\). Find the value of \(p\) and the value of \(q\).

METHOD 1

\(\alpha  + \beta  = 5,\,\,\alpha \beta  =  - 7\)     (M1)(A1)

Note: Award M1A0 if only one equation obtained.

\(\left( {\alpha  + 1} \right) + \left( {\beta  + 1} \right) = 5 + 2 = 7\)     A1

\(\left( {\alpha  + 1} \right)\left( {\beta  + 1} \right) = \alpha \beta  + \left( {\alpha  + \beta } \right) + 1\)     (M1)

\( =  - 7 + 5 + 1 =  - 1\)

\(p =  - 7,\,\,q =  - 1\)       A1A1

METHOD 2

\(\alpha  = \frac{{5 + \sqrt {53} }}{2} = 6.1 \ldots {\text{;}}\,\,\beta  = \frac{{5 - \sqrt {53} }}{2} =  - 1.1 \ldots \)     (M1)(A1)

\(\alpha  + 1 = \frac{{7 + \sqrt {53} }}{2} = 7.1 \ldots {\text{;}}\,\,\beta  + 1 = \frac{{7 - \sqrt {53} }}{2} =  - 0.1 \ldots \)     A1

\(\left( {x - 7.14 \ldots } \right)\left( {x + 0.14 \ldots } \right) = {x^2} - 7x - 1\)     (M1)

\(p =  - 7,\,\,q =  - 1\)       A1A1

Note: Exact answers only.

[6 marks]