Mathematics - Quadratic Equation Question with Solution | TestHub

The expression is $P(x)=(m^2-1)x^2+2(m-1)x+1$. A. For $P(x)>0$ for all $x\in \mathbb{R}$: If $m=1$, $P(x)=1$, which is always positive. If $m^2-1\ne 0$, we need $m^2-1>0⟹m\in (-\infty ,-1)\cup (1,\infty )$ and discriminant $D<0$. $D=(2(m-1){)}^2-4(m^2-1)(1)=4(m-1)[(m-1)-(m+1)]=-8(m-1)$. $D<0⟹-8(m-1)<0⟹m>1$. Combining these, $m>1$. Including $m=1$, $P(x)>0$ for $m\ge 1$. This statement is correct. B. For $P(x)<0$ for some $x\in \mathbb{R}$: If $m^2-1<0⟹m\in (-1,1)$, the parabola opens downwards, so $P(x)$ will take negative values. If $m=-1$, $P(x)=-4x+1$, which is negative for $x>1/4$. Thus, if $m\in (-1,1)$, $P(x)$ can be negative. This statement is correct. C. The equation $P(x)=0$ has real roots: If $m=1$, $P(x)=1$, so $1=0$ has no real roots. If $m=-1$, $P(x)=-4x+1=0$ has $x=1/4$ as a real root. If $m^2-1\ne 0$, for real roots, $D\ge 0⟹-8(m-1)\ge 0⟹m\le 1$. Combining, $P(x)=0$ has real roots if $m<1$ or $m=-1$. The statement "if $m\le 1$ " includes $m=1$, for which there are no roots. Thus, this statement is incorrect. D. If $m=-1$, $P(x)=((-1{)}^2-1)x^2+2(-1-1)x+1=-4x+1$. The equation $-4x+1=0$ yields $x=1/4$, which is a unique real root. This statement is correct.