solution set?

Question

I have to find the solution set of

x² - 3x < 4

I could really use some help.

Answer 1

So x^2-3x-4 < 0, or (x-4)(x+1) < 0. Then Either x>4 AND x < -1, or x<4 AND >-1. Obviously the first case is impossible (can you see why?), so we have -1
Steve

Answer 2

This algebraic formula is a theorem, in that it starts out with the supposition that all squared numbers, when subtracted to that number multiplied by three, would give a difference less than four.

Suppose that X=1:

[(1X1) - (3X1)] < 4
= [(1) - (3)] < 4
= [ -2] < 4.



Or if X = 3:

[(3X2) - (3X3)] < 4
= [(6) - (9)] < 4
= -3 < 4


To prove the theorem, just substitute X with any number, multiply it by itself (to get the square), subtract it from the product of that number and three, and then the result is less than four.
-- Graciana

Answer 3

x^2-3x-4<0
(x-4)(x+1)<0
x must be inbetween -4 and 1
x<-4 x>1
-4>x>1

Answer 4

x² - 3x < 4
x² - 3x - 4 < 0
-3² - 4.1.-4
9 + 16 = 25

x' = [-(-3) + \/25)] : 2
x' = [3 + 5] : 2 = 4

x" = [-(-3) - \/25)] : 2
x" = [3 - 5] : 2 = -2 : 2 = -1

Solution: {-1 .::.

Answer 5

x² -3x < 4

x² - 3x - 4 < 4 - 4

x² - 3x - 4 < 0

( x - 4)(x + 1) < 0

- - - - - - -

Roots

X = 4

x = - 1

- - - - - - - -s-

Answer 6

i got the answer for you:

(x-4)(x+1)

then x must either be 4 or -1