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# Range of values which are real? watch

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1. Find the range of values of p for which the roots of the equation px^2 + px -2 = 0

my working:

b^2 - 4ac

p2 -4p*-2
p2 + 8p >= 0
p (p + 8) >= 0
p >= -8
p >= 0

the correct answer is: p <= -8

where did i go wrong?
2. Where did you get p>=-8 from?
3. (Original post by syNK)
Find the range of values of p for which the roots of the equation px^2 + px -2 = 0

my working:

b^2 - 4ac

p2 -4p*-2
p2 + 8p >= 0
p (p + 8) >= 0
p >= -8
p >= 0

the correct answer is: p <= -8

where did i go wrong?

-1 * -2 >= 0, therefore -1 >= 0 and -2 >= 0
4. i think the equation is:
discriminant := p^2-4(p)(-2)>=0
simplifies to p(p+8)>=0

visualise the quadratic graph, so p>=0 or p<=-8 because y is +ve there
5. Why is it, p <=-8 instead of p >= -8 , why did the signs changes?
6. (Original post by syNK)
Why is it, p <=-8 instead of p >= -8 , why did the signs changes?
Why do you expect the signs to stay the same? Like I said above, -1 * -2 >= 0, but neither of them are >= 0 themselves.

Similarly, -1*5 =< 0, but -1 =< 0 and 5 >= 0
7. To solve the quadratic inequation, you have to draw the graph of and then determine when from the graph.
8. It is like the other quadratic that we did with you ... find the points it crosses the x-axis (-8 and 0) and then consider wether you want outside of these or between them
9. because when the expression p(p+8) is +ve, both "bits" - the p and the (p+8) have to be +ve simultaneously - which means that p>=0 AND p>=-8, which only happens if p>=0!

Likewise, p(p+8) is also +ve if both are -ve, using the same argument.

i find it useful, in each bit, to draw a number line, and see where both "bits" of the inequality intersect, which give you the values you want
10. (Original post by syNK)
Why is it, p <=-8 instead of p >= -8 , why did the signs changes?
draw the graph and think..
11. (Original post by Hasufel)
because when the expression p(p+8) is +ve, both "bits" - the p and the (p+8) have to be +ve simultaneously - which means that p>=0 AND p>=-8, which only happens if p>=0!

Likewise, p(p+8) is also +ve if both are -ve, using the same argument.

i find it useful, in each bit, to draw a number line, and see where both "bits" of the inequality intersect, which give you the values you want
what? why not just draw the graph and then you can see when ?
12. because - if you ever want to do engineering or mathematics or physics at university and beyond, and you have a problem in, say, differential equations which can only be solved by numerical methods, drawing a graph isn`t enough. You have to be able to prove...something...mathematical ly.
13. I'd draw a graph and I'm doing 3rd year maths.

EDIT: also drawing a graph would show you where you've gone wrong
14. Yes, but what if your inequality was [(x-1)(x-2)(x+4)]/[(x+1)(x-5)] ?
15. You can still draw the graph of that pretty easily (you just have to know where the asymptotes and zeros are).
16. I take your point - it`s all about speed in a classroom i suppose!

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Updated: September 29, 2011
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