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Consider the function $f(x)=mx^2+(2m-1)x+(m-2)$. Choose the correct options:

$(A).$ If $f(x)>0$ for all $x \in R$, then $m \in (- \infty, -1/4)$

$(B).$ The number of integral values of $m$ greater than $-5$ so that $f(x)<0$ for all $x \in R$ are 4.

$(C).$ The number of integral values of $m$ less than $50$ so that roots of the quadratic equation $f(x)=0$ are rational are 6.

$(D)$ The curve $y=f(x)$ touches the $x-axis$ if $m=-1/4$

I have found that option($2,4$) are correct and $1$ is wrong. I need help with option $3$. I know for rational roots Discriminant should be square of a rational number and coefficients should be rational too but not able to apply it here.

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  • $\begingroup$ Option A seems to be vacuously true, since "$f(x)>0$ for all $x$" is false for any $m$. Or am I missing something? $\endgroup$ – Arthur Jul 24 '16 at 9:40
  • $\begingroup$ Maybe I'm wrong but the way I see the (D) option is to verify that for $m=- \frac 1 4$ the curve touches the x-axis $\endgroup$ – user261263 Jul 24 '16 at 9:41
  • $\begingroup$ @Arthur Actually it was $-1/4$. 'Typo' $\endgroup$ – MathGeek Jul 24 '16 at 9:42
  • $\begingroup$ Doesn't matter. I still think the statement is vacuously true. $\endgroup$ – Arthur Jul 24 '16 at 9:43
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Hint:

the discriminant of original polynomial can be written as $(2k+1)^2 = 4(k^2+k)+1$

EDIT: If one still can not understand one can expand the discriminant and get $4m+1$ and then

$4m+1 = 4(k^2+k)+1$ which means $m = k^2+k\leq 50 \implies 1 \leq k \leq 6$. There are 6 rational roots.

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They ask for integer values of $m$, so rational coefficients are taken care of automatically. The discriminant is $$ (2m-1)^2-4m(m-2)=4m+1 $$ which is supposed to be a square of a rational number. Note that as long as $m$ is limited to integers, we might as well ask for perfect squares.

So, how many values of $m$ below $50$ makes $m$ a square? Clearly, said square cannot be even. However, if we have an odd square $(2n+1)^2$, then $$ (2n+1)^2=4n^2+4n+1=4(n^2+n)+1 $$ is of that form. Since the largest possible value of $4m+1$ is $201=15^2-24=14^2+5$, there are seven possible odd squares: $1^2,3^2,5^2,7^2,9^2,11^2$ and $13^2$.

However, $4m+1=1^2$ corresponds to $m=0$. Technically it still fulfills the requirement that "all roots are rational", but the equation is not quadratic anymore, so whether or not it counts is up to whoever wrote the problems in the first place. I'd guess it doesn't count, which means option 3 is true.

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