How to solve this inequality.?

Question

y^4+y^3 < 4(y+4)

I know you have to first distribute the 4 to get

y^4+y^3 < 4y+16

Then maybe set it to 0:

y^4+y^3 -4y - 16 < 0

But then what? I thought maybe I can take a y out and a 4 out...

y^3(y+1) -4(y+4) < 0

But then I'm stuck. Can someone please help?

I also have another question....

2x^2+7x+8

---------------- > 0

x^2 + 1

If I do the quadratic equation on the top, I get a negative in the square root.. but I'm not sure if that is a good thing. Can someone help me on this problem as well?

Thanks a lot!

Answer 1

Wouldn&#039;t you combine like term to get

11y - 16 &lt; 0

Isnt that all?

Oh it&#039;s been a while since I&#039;ve been in school.

Good Luck Hope I helped.

Answer 2

For your first question:

y^4 + y^3 &lt; 4(y+4)

y^4 + y^3 &lt; 4y + 16

y^4 + y^3 - 4y - 16 &lt; 0

-2 &lt; x &lt; 2

For your second question:

2x^2 + 7x + 8 &gt; 0

All values of x result in a number great than 0.