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

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 $\displaystyle x\in\mathbb{R}$ .

Re: Finding for interval of m in quadratic equation.

Quote:

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???

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 $\displaystyle x_2=\frac{3-\sqrt{9+16m}}{8}$ .

Re: Finding for interval of m in quadratic equation.

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

Re: Finding for interval of m in quadratic equation.

Quote:

Originally Posted by

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

$\displaystyle m<-9/16$ .

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.