Question

The number of real roots of the equation |x^2+x-6| + 2|x| - 4 =0 is?

Answer 1

@satellite73 @Hero @hartnn

Answer 2

I'm guessing six right now. I'll have to play with it a little more to be sure.

Answer 3

U see @CliffSedge " REAL ROOTS"

Answer 4

I know, I know, that's just a first approximation to set some bounds . . .

Answer 5

The Options are 0 1 2 3

Answer 6

Hmm, must have done something wrong, because now I am getting 8 solutions. I'll have to check them with the original equation to see if some are erroneous.

Answer 7

if i use trial error, i get 2

Answer 8

Hint: Use the rule \[|a| =\sqrt{ a^2}\] to solve

Answer 9

I know x=2 works by inspection.
I only found one other that works.
I'll also say only 2 solutions.

Answer 10

The answer shuld be 1 solution.....)

Answer 11

\[|a| = \pm a\]

Answer 12

|x^2+x-6| + 2|x| - 4 =0

x^2 + x - 6 + 2x - 4 =0

x^3 + 3x - 10 =0 --------- 1


|x^2+x-6| + 2|x| - 4 =0


-x^2-x+6 - 2x - 4 =0

x^2 + 3x - 2 =0 ----------2

Answer 13

for x>0, 1st equation should be :
x^2 + 3x - 10 = 0
(x+5)(x-2) = 0
x=2 is which satisfy be a solution

Answer 14

I see where I made an error (lost track of a negative sign . . .).
Now I only see x=2 as solution.

Answer 15

if u substitute back the other 3 roots u got in original equation, it will not get satisfied. so x=2 is the only real solution