Need help: Find for the values....

**Ah Chan** · August 25th 2009, 07:08 PM

Hi, everyone..I'm a new recruit and just joined this forum several weeks ago. So, hope u all can help me through every difficulties I face..yeah, this is my first ever question and I hope there's somebody can help me out...I'm not sure its the question problems or its mine problems..anyways have a look, thanks..

a) Find the set of values of m for which the graph of y = (3 - m)x^2 + 4x - m lies above the x-axis for all real values of x.

**mathaddict** · August 25th 2009, 07:31 PM

Originally Posted by Ah Chan

Hi, everyone..I'm a new recruit and just joined this forum several weeks ago. So, hope u all can help me through every difficulties I face..yeah, this is my first ever question and I hope there's somebody can help me out...I'm not sure its the question problems or its mine problems..anyways have a look, thanks..

a) Find the set of values of m for which the graph of y = (3 - m)x^2 + 4x - m lies above the x-axis for all real values of x.

Hello fellow Malaysian ,

note that it has no real roots since it is floating above the x-axis and this implies that its discriminant ie $b^2-4ac<0$

where a=3-m , b=4 and c=-m .

Then this is a positive graph , where a>0 , so you will need to pick the correct ranges of m at the end .

**Soroban** · August 25th 2009, 08:07 PM

Hello, Ah Chan!

Find the set of values of $m$ for which the graph of . $y \:=\: (3 - m)x^2 + 4x - m$
lies above the $x$ -axis for all real values of $x.$

First of all, the parabola must open upward.
. . Hence, the leading coefficient must be positive: . $3-m \:>\:0 \quad\Rightarrow\quad m \:<\:3$ .[1]

Second, there are no $x$ -intercepts.
. . That is, the equation: . $(3-m)x^2 + 4x - m \:=\:0$ . has no real roots.

Quadratic Formula: . $x \;=\;\frac{-4 \pm\sqrt{4^2 - 4(3-m)(-m)}}{2(3-m)} \;=\;\frac{-2 \pm\sqrt{(1+m)(4-m)}}{3-m}$

To have no real roots, the discriminant must be negative: . $(1+m)(4-m) \:<\:0$

There are two cases: . $\begin{array}{ccc}(1) & \text{(pos)}\times\text{(neg.)} \\ \\[-4mm] (2) & \text{(neg)}\times\text{(pos)} \end{array}$

$(1)\;\begin{array}{ccc}1+m \:>\:0 & \Longrightarrow & m \:>\:\text{-}1 \\ 4-m \:<\:0 & \Longrightarrow & m \:>\:4 \end{array} \quad\Longrightarrow\quad m > 4\quad\hdots$ . which contradicts [1].

$(2)\:\begin{array}{ccc}1+m \:<\:0 & \Longrightarrow & m \:<\:\text{-}1 \\ 4-m \:>\:0 & \Longrightarrow & m \:<\:4\end{array}\quad\Longrightarrow\quad m \:<\:-1 \quad\hdots$ . which agrees with [1].

. . . . Therefore: . $m \;<\;-1$

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