Let $f\left( x \right) = p{x^2} + qx - 4p$, where p ≠ 0. Find Find the number of roots for the equation $f\left( x \right) = 0$.

Justify your answer.

METHOD 1

evidence of discriminant      (M1)
eg  ${b^2} - 4ac,\,\,\Delta$

correct substitution into discriminant      (A1)
eg  ${q^2} - 4p\left( { - 4p} \right)$

correct discriminant       A1
eg  ${q^2} + 16{p^2}$

$16{p^2} > 0\,\,\,\,\left( {{\text{accept}}\,\,{p^2} > 0} \right)$     A1

${q^2} \geqslant 0\,\,\,\,\left( {{\text{do not accept}}\,\,{q^2} > 0} \right)$     A1

${q^2} + 16{p^2} > 0$      A1

$f$ has 2 roots     A1 N0

METHOD 2

y-intercept = −4p (seen anywhere)      A1

if p is positive, then the y-intercept will be negative      A1

an upward-opening parabola with a negative y-intercept      R1
eg  sketch that must indicate p > 0.

if p is negative, then the y-intercept will be positive      A1

a downward-opening parabola with a positive y-intercept      R1
eg  sketch that must indicate p > 0.

$f$ has 2 roots     A2 N0

[7 marks]