Question

If the quadratic equation $x^2-mx-4x+1=0$ has real and distinct roots, then the values of $m$ are
A. $(-\infty ,-6)$
B. $(-\infty ,-3)$

C. $(-2,\infty )$

D. $(2,\infty )$

• A. A or C
• B. A or D
• C. B or C
• D. B or D

Answer 1

Answer: (A)

A or C

For an equation $a{x}^2+bx+c=0$, the discriminant $△=b^2-4ac$ helps us understand the nature of the roots.
When $△>0$ and a perfect square, roots are real and distinct
So ${(-(m+4\left)\right)}^2-4\times 1\times 1>0$

$\Rightarrow$ $(m^2+16+8m)-4>0$

$\Rightarrow$$m^2+8m+12>0$

$\Rightarrow$$m^2+2m+6m+12>0$

$\Rightarrow$$m(m+2)+6(m+2)>0$

$\Rightarrow$$(m+6)(m+2)>0$
When both $(m+6)>0;(m+2)>0$

$\Rightarrow$$m>-6;m>-2$

From these two solutions, we have $m>-2$
Also, $(m+6)(m+2)>0$ , when both $(m+6)<0;(m+2)<0$

$\Rightarrow$$m<-6;m<-2$

From these two solutions, we have $m<-6$
So both $A$ and $C$ are the possible values of $m$