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

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

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

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

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

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

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

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

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

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

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

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 m

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