Question

If equation $R\left(x\right)=x^2+mx+1$ has two distinct real roots, then exhaustive values of $m$ are :

• A. $(-2,2)$
• B. $(-\infty ,-2)\cup (2,\infty )$
• C. $(-2,\infty )$
• D. all real number

Answer 1

Answer: (D)

all real number

Let unknown polynomial be $P\left(x\right)$.
Let $Q\left(x\right)$ and $R\left(x\right)$ be the quotient and remainder respectively, when it is divided by $(x-3)(x-4)$. Then,
$P\left(x\right)=(x-3)(x-4)Q\left(x\right)+R\left(x\right)$
Then, we have
$R\left(x\right)=ax+b$
$\Rightarrow P\left(x\right)=(x-3)(x-4)Q\left(x\right)+ax+b$
Given that $P\left(3\right)=2$ and $P\left(4\right)=1$. Hence,
$3a+b=2$ and $4a+b=1$
$\Rightarrow a=-1$ and $b=5$
$\Rightarrow R\left(x\right)=5-x$
$5-x=x^2+ax+1\Rightarrow x^2+(m+1)x-4=0$
Given that roots are real and distinct. Therefore,
$D>0\Rightarrow (m+1{)}^2+16>0$
Which is true for all real $m$.

Hence, option D is correct.