1. ## Finding for interval of m in quadratic equation.

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.

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 hint or steps to solve the problem

2. ## Re: Finding for interval of m in quadratic equation.

For all $x,m\in \mathbb{R}$ , numerator and denominator belong to $\mathbb{R}$ . You only need to study for which values of $m$ there exists $x$ such that $-4x^2+3x+m=0$ . For those values of $m$ the quotient is not defined for all $x\in\mathbb{R}$ .

3. ## Re: Finding for interval of m in quadratic equation.

Originally Posted by FernandoRevilla
. You only need to study for which values of $m$ there exists $x$ such that $-4x^2+3x+m=0$ .

how to find m
should i equate D=0 or -D/4a=0???

4. ## Re: Finding for interval of m in quadratic equation.

The roots of $-4x^2+3x+m=0$ are $x_1=\frac{3+\sqrt{9+16m}}{8}$ and $x_2=\frac{3-\sqrt{9+16m}}{8}$ .

5. ## Re: Finding for interval of m in quadratic equation.

the final answer is m $\epsilon$[1,7]

6. ## Re: Finding for interval of m in quadratic equation.

Originally Posted by sumedh
the final answer is m $\epsilon$[1,7]
$m<-9/16$ .

7. ## Re: Finding for interval of m in quadratic equation.

Then the question might be wrong

Thank you very much for your help and the valuable time you spared for me.

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### find the interval in which m lies so that the expression mx² 3x-4

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