# Finding for interval of m in quadratic equation.

• Aug 23rd 2011, 10:44 PM
sumedh
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
• Aug 23rd 2011, 11:02 PM
FernandoRevilla
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 $\displaystyle x\in\mathbb{R}$ .
• Aug 24th 2011, 12:22 AM
sumedh
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???
• Aug 24th 2011, 12:36 AM
FernandoRevilla
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 $\displaystyle x_2=\frac{3-\sqrt{9+16m}}{8}$ .
• Aug 24th 2011, 02:08 AM
sumedh
Re: Finding for interval of m in quadratic equation.
the final answer is m$\displaystyle \epsilon$[1,7]
• Aug 24th 2011, 06:57 AM
FernandoRevilla
Re: Finding for interval of m in quadratic equation.
Quote:

Originally Posted by sumedh
the final answer is m$\displaystyle \epsilon$[1,7]

$\displaystyle m<-9/16$ .
• Aug 25th 2011, 10:25 AM
sumedh
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.