Answer to Question #233060 in Algebra for dizzo

Question

Discuss the roots and signs of the equation : (m+1)x^2+4(m-1)x+4-m=0

Answer 1

We have to use the general equation to solve the second-degree formula and determine a, b and c:

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

We substitute and find x: $x=\frac{-b\plusmn \sqrt{b^2-4a\cdot c}}{2a}$

$x=\dfrac{-4(m-1)\plusmn \sqrt{[4(m-1)]^2-4(m+1) (4-m)}}{2(m+1)} \\ x=\dfrac{4-4m\plusmn \sqrt{[16(m^2-2m+1)]-4(4+3m-m^2)}}{2m+2} \\ x=\dfrac{4-4m\plusmn \sqrt{16m^2-32m+16-16-12m+4m^2)}}{2m+2} \\ x=\dfrac{4-4m\plusmn \sqrt{20m^2-44m}}{2m+2}=\dfrac{4-4m\plusmn \sqrt{4(5m^2-11m)}}{2m+2} \\ x=\dfrac{2(2-2m)\plusmn 2\sqrt{5m^2-11m}}{2(m+1)}=\dfrac{2-2m\plusmn \sqrt{5m^2-11m}}{m+1} \\ x=\dfrac{2-2m\plusmn \sqrt{(5m-11)m}}{m+1}$

Then, we can observe that the function has the following critical points $m={\{ -1, 0, \frac{11}{5} \} }$.

When we consider $m<-1$ to find the values of x we will find two possible real values for x (one positive and one negative). On the next interval $-1<m\le 0$ we have two possible positive values for x. The following interval $0<m<\frac{11}{5}$ gives an imaginary solution for x in the form a+bi. The values following $\frac{11}{5}<m<4$ give two possible negative values for x, while for $m\ge 4$ we will have two possible values for x (one positive and one negative).