Show that $(m+1)x^2+(4m+3)x+2m=0$ has real and distinct roots for all real values of $m$.[4

Question

Show that $(m+1)x^2+(4m+3)x+2m=0$ has real and distinct roots for all real values of $m$.[4] The equation of a curve is $y=4x^2-4x+3$. Find the set of values of $x$ for which the curve lies below the line $y=11$.[3] The straight line $L$ meets the curve $y=4x^2-4x+3$ at one point only. Given that $L$ is not a tangent to the curve, what can be deduced about $L$?[1] Correction to part (a): the two-root statement requires $m\ne-1$. For $m=-1$, the equation is linear. Address this exceptional case separately.

Accepted Answer

(a) For $m\ne-1$, the equation has two real and distinct roots. For $m=-1$, $x=-2$ is its only root, so the printed claim is false. (b)(i) $-1<x<2$. (b)(ii) $L$ is parallel to the $y$-axis.

2025 TKSS PRELIMS P2 Q8

2025 TKSS PRELIMS P2 Q8

Solution:

Solution locked