Question

Given the equation in X, (m + 1)X^2+ 4(m – 1)X + 4 – m = 0, determine m in each of

the following cases:

a) The equation has a double root, what is the root? (3marks)

b) the equation is of the first degree. What is the root? (3marks)

c) 0 is a root of the equation. What is the other root? (3marks)

d) the equation has two roots of different signs (3marks)

e) the equation has two positive roots. (3marks)

Answer 1

$(m+1)x^2+4(m-1)x+4-m=0$

a) The equation has the double root if $D=4^2(m-1)^2-4(m+1)(4-m)=20m^2-44m=20m(m-2.2)=0$ .

If $m=0$ , then we have $x^2-4x+4=(x-2)^2=0$ . The root is $x=2$ .

If $m=2.2$ , then we have $3.2x^2+4.8x-1.8=3.2(x-0.75)^2=0$ . The root is $x=0.75$

b) If the equation is of the first degree, then $m+1=0$ and $m=-1$ .

We have the following equation: $-8x+5=0$

The root is $x=5/8=0.625$

c) If 0 is the root of the equation, then $4-m=0$ and $m=4$ .

We have the following equation: $5x^2+12x=0$

$5x^2+12x=5x(x+2.4)=0$

The other root is $x=-2.4$

d) The equation has two roots of different signs, if $D>0$ (existence) and $x_1x_2<0$ (different signs).

$\begin{cases} D=22m(m-2.2)>0 \\ x_1x_2=\frac{c}{a}=\frac{4-m}{m+1}<0 \end{cases}$

$\begin{cases} m\in (-\infty , \ 0)\cup (2.2,\ +\infty) \\ m\in (-\infty,\ -1)\cup (4,\ +\infty) \end{cases}$

$m\in (-\infty,\ -1)\cup (4,\ +\infty)$

e) The equation has two (distinct) postings roots, if $D>0$ (existence) and $x_1x_2>0$ , $x_1+x_2>0$ (positive).

$\begin{cases} D=22m(m-2.2)>0 \\ x_1x_2=\frac{c}{a}=\frac{4-m}{m+1}>0 \\ x_1+x_2=-\frac{b}{a}=-\frac{4(m-1)}{m+1}>0 \end{cases}$

$\begin{cases} m\in (-\infty , \ 0)\cup (2.2,\ +\infty) \\ m\in (-1,\ 4)\\ m\in (-1,\ 1) \end{cases}$

$m\in(-1,0)$

Answers:

a) $m=0,\ x=2$ and $m=2.2, \ x=0.75$

b) $m=-1,\ x=0.625$

c) $m=4,\ x=-2.4$

d) $m\in (-\infty,\ -1)\cup (4,\ +\infty)$

e) $m\in(-1,0)$