solve inequation x^2 + 4ax < 0

[acemi123]

solve inequation with 2 unknown! What is the way to solve this type of inequations?
x^2+4ax<0

a-) 0<x<4a for every a Real number
b-) no solution if a>0
c-) -4a<x<0 if a>0
d-) -4a<x<0 if a<0

Thanks.

 

[Harry_the_cat]

"a" is considered a constant. x is the variable.

Note that x² + 4ax = x(x +4a).

I like solving quadratic inequalities by considering the graph. There are also more algebraic ways.

What would the graph of y = x² + 4ax look like?
When will this graph be below the x-axis?

 

[JeffM]

solve inequation with 2 unknown! What is the way to solve this type of inequations?
x^2+4ax<0

a-) 0<x<4a for every a Real number
b-) no solution if a>0
c-) -4a<x<0 if a>0
d-) -4a<x<0 if a<0

Thanks.

Be systematic.

Solve the corresponding equality first.

[MATH]x^2 + 4ax = 0 \implies x(x + 4a) = 0 \implies x = 0 \text { or } x = - \ 4a.[/MATH]
So you know that

[MATH]x = 0 \implies x^2 + 4ax \not < 0 \text {, and } x = -\ 4a \not < 0.[/MATH]
So you have four cases to to consider, namely:

(1) x > 0 and x > - 4a, (2) x > 0 and x < - 4a, (3) x < 0 and x > - 4a, and (4) x < 0 and x < - 4a.

Can you figure it out now?

 

[HallsofIvy]

The answer, of course, depends on a. In particular whether a is positive or negative.

x^2- 4ax= x(x- 4a)$. In order that the product of two number be negative, one must be positive and the other negative so either
(i) x< 0, x- 4a> 0 or
(ii) x> 0, x- 4a< 0.

In case (i) x> 4a so that x< 0 can only be true if 4a is negative. If a is negative then 4a< x< 0.

In case (ii) x< 4a so that x> 0 can only be true if 4a is positive. If a is positive then 0< x< 4a.