Thread: Quadratic inequalities

1. Quadratic inequalities

Two inequality questions (not related)

1. if k does not equal 0 solve the quadratic inequation for all real values
k^2x+kx+2k<0
2. Consider the quadratic equation ax^2+px+aq+q=0 where, a does not equal 0 & p and q are constants.
It is known that one of the roots of the quadratic equation is always 1 regardless of the value of a. Prove that p+q=0

2. Originally Posted by requal
Two inequality questions (not related)

1. if k does not equal 0 solve the quadratic inequation for all real values
k^2x+kx+2k<0

Mr F asks: What pronumeral is being solved for .... x or k?

2. Consider the quadratic equation ax^2+px+aq+q=0 where, a does not equal 0 & p and q are constants.
It is known that one of the roots of the quadratic equation is always 1 regardless of the value of a. Prove that p+q=0
2. If x = 1 is always a root then a + p + aq + p = 0 => p + q + a(1 + q) = 0. Since $a \neq 0$ it follows that if p + q = 0 then q = -1. So it seems to me that the statement is only true if q = -1.

3. doesnt say but i think its referring to x though i guess if u solve one of them you will know the other

4. Originally Posted by requal
doesnt say but i think its referring to x though i guess if u solve one of them you will know the other
I too thought it might be x except that it's linear, not quadratic in x ....

5. yep my bad; it is actually a typo; its supposed to be k(x^2)+kx+2k<0
i think is the second time this has happened. No wonder it takes so long for people to respond to my posts

6. Originally Posted by requal
yep my bad; it is actually a typo; its supposed to be k(x^2)+kx+2k<0
i think is the second time this has happened. No wonder it takes so long for people to respond to my posts
Hi requal,

In that case, you could just divide by k both side since k is not zero.
Then you are just working with $x^2+x+2 < 0$

You may want to check the signs though, it seems like there is no real solution...

Edit:
My bad, if k > 0, the answer is "no real solution" like shown above.
if k < 0, then the "<" switches ">" when divided by k, which would mean the answer is all real numbers.