SOLUTION: I was womdering if you could explain how to answer 3 problems:
1) x^2+x>0
2) x+3<-2/x
3) y=x^2+4x+1
Please explain each step so that I can use these are sample and mode
Algebra ->
Quadratic Equations and Parabolas
-> SOLUTION: I was womdering if you could explain how to answer 3 problems:
1) x^2+x>0
2) x+3<-2/x
3) y=x^2+4x+1
Please explain each step so that I can use these are sample and mode
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Question 5429: I was womdering if you could explain how to answer 3 problems:
1) x^2+x>0
2) x+3<-2/x
3) y=x^2+4x+1
Please explain each step so that I can use these are sample and model problems.
Thanks,
Char Answer by longjonsilver(2297) (Show Source):
then we can say that either x=0 or x+1=0, so answers are x=0 or x=-1.
2. x+3<-2/x --> get rid of the fraction by multiplying every term by x, to give or . First lets, find where --> we again factorise, to give (x+1)(x+2) = 0. So x=-1 or x=-2.
So going back to the question, , where is the curve less than zero: we have found where the curve crosses the x-axis (at x=-2 and x=-1) and the curve is a u=shape, so the curve has -ve y-values between the x=-2 and x=-1 values: plot it to see this.
So answer is -2
3. . This does not factorise, so you will have to use the quadratic formula to work out the x-values. You have a go at this.
