SOLUTION: I need help with my sons Algebra II, it has been many years... Solve the inequality algebraically x 2 -3x-4 < 0 (that is x squared)

Algebra ->  Algebra  -> Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: I need help with my sons Algebra II, it has been many years... Solve the inequality algebraically x 2 -3x-4 < 0 (that is x squared)      Log On

Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations!
Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help!

   

Question 113449: I need help with my sons Algebra II, it has been many years...
Solve the inequality algebraically

x 2 -3x-4 < 0 (that is x squared)
Found 2 solutions by MathLover1, josmiceli:
Answer by MathLover1(3376) About Me  (Show Source):
You can put this solution on YOUR website!

x+%5E2+-3x-4+%3C+0
We need to solve it as an equation and find roots.
x+%5E2+-3x-4+=+0

Here is the solution with graph:

Solved by pluggable solver: Quadratic Formula
Let's use the quadratic formula to solve for x:


Starting with the general quadratic


ax%5E2%2Bbx%2Bc=0


the general solution using the quadratic equation is:


x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29




So lets solve x%5E2-3%2Ax-4=0 ( notice a=1, b=-3, and c=-4)





x+=+%28--3+%2B-+sqrt%28+%28-3%29%5E2-4%2A1%2A-4+%29%29%2F%282%2A1%29 Plug in a=1, b=-3, and c=-4




x+=+%283+%2B-+sqrt%28+%28-3%29%5E2-4%2A1%2A-4+%29%29%2F%282%2A1%29 Negate -3 to get 3




x+=+%283+%2B-+sqrt%28+9-4%2A1%2A-4+%29%29%2F%282%2A1%29 Square -3 to get 9 (note: remember when you square -3, you must square the negative as well. This is because %28-3%29%5E2=-3%2A-3=9.)




x+=+%283+%2B-+sqrt%28+9%2B16+%29%29%2F%282%2A1%29 Multiply -4%2A-4%2A1 to get 16




x+=+%283+%2B-+sqrt%28+25+%29%29%2F%282%2A1%29 Combine like terms in the radicand (everything under the square root)




x+=+%283+%2B-+5%29%2F%282%2A1%29 Simplify the square root (note: If you need help with simplifying the square root, check out this solver)




x+=+%283+%2B-+5%29%2F2 Multiply 2 and 1 to get 2


So now the expression breaks down into two parts


x+=+%283+%2B+5%29%2F2 or x+=+%283+-+5%29%2F2


Lets look at the first part:


x=%283+%2B+5%29%2F2


x=8%2F2 Add the terms in the numerator

x=4 Divide


So one answer is

x=4




Now lets look at the second part:


x=%283+-+5%29%2F2


x=-2%2F2 Subtract the terms in the numerator

x=-1 Divide


So another answer is

x=-1


So our solutions are:

x=4 or x=-1




As you can see, your solutions are: x%5B1%5D+=+4 and x%5B2%5D+=+-1
For these values x+%5E2+-3x-4 will be equal to +0. That means for all values of x from -1 (excluding -1) to 4 (excluding 4), you will have x+%5E2+-3x-4 less than 0.
or x+%5E2+-3x-4+%3C+0 for -1%3C+x+%3C+4
As you can see on graph, that part of the parabola lies in III-th and IV-th quadrant (under x-axis)



Answer by josmiceli(6784) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E2+-+3x+-+4+%3C+0
%28x+-+4%29%28x+%2B+1%29+%3C+0
OK, I've got two things multiplied together giving me less than zero
which means negative. This is like
a%2Ab+%3C+0
Here are the possibilities:
(+)*(+) = +
(-)*(-) = +
(+)*(-) = -
(-)*(+) = -
I don't want the 1st two cases, only the 2nd two.
How do I make (x + 1) negative? By making x less than -1
x+%3C+-1
But if x+%3C+-1, (x - 4) ends up being negative, too and I get
the case (-)*(-) = +, and I don't want that
So, I'll choose x so that (x - 4) is negative
x+%3C+4 makes it negative, but I want (x + 1) to stay positive, too
(x + 1) stays positive when x+%3E+-1
Now I combine these
-1+%3C+x+%3C+4
Notice I only can use the case (-)*(+) = -
Hope you can follow this & hope I got it right, too