Finding for interval of m in quadratic equation.

[Sumedh]

Maths Quadratic Question

Homework Statement

Find the interval in which 'm' lies so that the expression
$$\frac{mx^2+3x-4}{-4x^2+3x+m}$$
can take all real values ,where x is real.

The Attempt at a Solution

i have equated this equation to y
$$y=\frac{mx^2+3x-4}{-4x^2+3x+m}$$

then

$$(-4y-m)x^2+(3y-3)x+(ym-4)=0$$


but then it becomes quadratic equation in many variables .


please give hints or steps to solve the problem.

 

[HallsofIvy]

One value of m for which y cannot "take all real values" is -4. Do you see why? Think about that.

 

[Sumedh]

i only have the answer and not the solution

and the answer is:-
$$m \epsilon [1,7]$$


but
as denominator cannot be zero

$$-4x^2+3x+m=0$$
the value of m in above equation will not be included in the interval

i took its discriminant
$$D =3^2-4(-4m)=9+16m$$
then
$$D\ge0$$
$$9+16m\ge0$$
$$m\ge-9/16$$

but this is not the answer?

 

[eumyang]

i took its discriminant
$$D =3^2-4(-4m)=9+16m$$
then
$$D\ge0$$

Again, you're confusing the idea of "having real solutions" with "taking on real values." If you want to say that the denominator cannot be zero, you are saying that, no matter what real-number value you plug in for x, the denominator won't be zero. D ≥ 0 means that there exists one or two values of x that make the quadratic equal to zero. That's not what you want here.

 

[Sumedh]

1) To get real value of the expression, the denominator should not be zero.
Is this statement correct??


2) As you said---------
D ≥ 0 means that there exists one or two values of x that make the quadratic equal to zero.
--------------
Should i use $$D\le0$$
if we use this we get $$m\le−9/16$$ (am i right till here?)

if i am right, then what should i do with $$m\le−9/16$$
as the final answer is
mϵ[1,7]