/sci/ - Science & Math

[Anonymous No.5684699]

Guys, this is a pilot question that I'm not supposed to know about, but I want to know.

I read that b^2-4ac is used here,
in my case
b^2-4ac > 0
so 2 real solutions... how do I find where j has no real solutions?
are there any points where there are no real solutions?
how do I write this as an inequality?
Bumping with /sci/ images

[Anonymous No.5684706]

[Anonymous No.5684709]

[Anonymous No.5684712]

bumping like a phaggot

[Anonymous No.5684725]

Anytime 4ac is larger than b^2 you will have an imaginary solution. That is anytime b^2-4ac < 0
I hope you know what a and c are.

And your inequality should be greater than or equal.

[Anonymous No.5684727]

When a quadratic polynomial is written in standard form (ax^2 + bx + c = f(x)), if b^2 - 4ac > 0 then your function has two real solutions. You are given a polynomial with two out of the three coefficients (see: a,b,c) defined. Substitute those values and simplify the inequality.

[Anonymous No.5684736]

help would be appreciated

[Anonymous No.5684742]

<span class="math">
49 > 24j
49 / 24 > j
[/spoiler]

[Anonymous No.5684749]

>>5684727
so I plug in a value for c so that I get a negative?
then I just move j to the other side of the inequality?
I think I got it, when j>49/24??

[Anonymous No.5684755]

lol wut
I don't understand thissss
forgive my stupidity, could you explain?

[Anonymous No.5684758]

>>5684742
>>5684727
so I can write the inequality as
j>49/24
is that correct?
because then the dicriminant would be less than 0 etc.

[Anonymous No.5684770]

>>5684749
if (72 - 4*6*j) is negative, the discriminant is negative, since you cannot take the root of a negative (in Real numbers), it will have no real solution. So that means that for (72 - 4*6*j) to be negative, j has to be smaller than (-72/(4*6)), so smaller than 2,041666. repeating

[Anonymous No.5684775]

>>5684770
fuck, thats supposed to be 7 to the power of 2 instead of 72

[Anonymous No.5684779]

>>5684775
gotcha!
thanks man!!