The equation $kx^{2}+(k+3)x+1=0$, where $k$ is a non-zero constant, is a quadratic equation in $x$.
(a) Show that the discriminant of the equation is $k^{2}+2k+9$. [2]
(b) Prove that, for all non-zero values of $k$, the equation has two distinct real roots. [3]
- (a) for attempting $b^{2}-4ac$ with $a=k,\ b=k+3,\ c=1$M1
- (a) $(k+3)^{2}-4k=k^{2}+6k+9-4k=k^{2}+2k+9$ (no errors – given answer)A1*
- (b) for completing the square: $k^{2}+2k+9=(k+1)^{2}+8$M1
- (b) for stating $(k+1)^{2}\geqslant 0$ so $(k+1)^{2}+8\geqslant 8>0$A1
- (b) for a full conclusion: discriminant $>0$ for all $k$, so two distinct real rootsA1
- Answer: (a) $k^{2}+2k+9$ (b) $(k+1)^{2}+8\geqslant 8>0$ so $b^{2}-4ac>0$ for all $k$
- Examiner tip: A proof needs a conclusion that links back: say 'discriminant $>0$, therefore two distinct real roots', not just '$>0$'.