1. ## Number and sign

m is the real parameter.
Find the real number and sign for the equation :

By two methods : graphic , algebric.

2. solve the equation in $\mathbb{C}$ perhaps ?

x^2+6x+9m=0

$\delta=36(1-m)$

for m>1 , m=1, m<1

3. Originally Posted by J.R
solve the equation in $\mathbb{C}$ perhaps ?

x^2+6x+9m=0

$\delta=36(1-m)$

for m>1 , m=1, m<1
Hello : Thank you this equation in R

4. so what happened when m<1 ?

m=1 ?

and m>1 ?

5. Graphic.

We have $m=-\frac{1}{9}x^2-\frac{2}{3}x$

Intersect the graph of $f(x)=-\frac{1}{9}x^2-\frac{2}{3}x$ with parallel lines to x-axis with equation $y=m$

If $m>1$ then the line doesn't intersect the graph.

If $m=1$ the line is tangent to the graph and the equation has a double solution $x=-3$

If $m\in(0,1)$ the line intersects the graph in 2 points and the equation has two negative solutions.

If $m=0$ then the equation has the solutions $x_1=-6, \ x_2=0$

If $m<0$ then the equation has one negative solution and one positive solution.

6. Originally Posted by red_dog
Graphic.

We have $m=-\frac{1}{9}x^2-\frac{2}{3}x$

Intersect the graph of $f(x)=-\frac{1}{9}x^2-\frac{2}{3}x$ with parallel lines to x-axis with equation $y=m$

If $m>1$ then the line doesn't intersect the graph.

If $m=1$ the line is tangent to the graph and the equation has a double solution $x=-3$

If $m\in(0,1)$ the line intersects the graph in 2 points and the equation has two negative solutions.

If $m=0$ then the equation has the solutions $x_1=-6, \ x_2=0$

If $m<0$ then the equation has one negative solution and one positive solution.
HELLO : ALGEBRIC SOLUTION
Student sign of : delta , p=x1 ×x2 and s=x1 + x 2