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# P1 - proof - HELP! watch

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1. hi can someone help me with this proof.

prove that the equation

x^2 + px + q = 0

has distinct, real roots. and only when p^2 > 4q

thank you if you can help

x
2. A quadratic has distinct real roots if the discriminant is greater than zero; in other words, if b^2 - 4ac > 0.
3. (Original post by rachio)
hi can someone help me with this proof.

prove that the equation

x^2 + px + q = 0

has distinct, real roots. and only when p^2 > 4q

thank you if you can help

x
p^2- 4q > 0

p^2 > 4q
4. (Original post by TheWolf)
p^2- 4q > 0

p^2 > 4q
is that all it is? the text book i have has some very very long complicated method which i do not understand. but if this is it it doesn't seem too hard.

thank you!
5. (Original post by rachio)
is that all it is? the text book i have has some very very long complicated method which i do not understand. but if this is it it doesn't seem too hard.

thank you!
b^2 - 4ac > 0 if there are real roots

so you just do it
6. (Original post by rachio)
is that all it is? the text book i have has some very very long complicated method which i do not understand. but if this is it it doesn't seem too hard.

thank you!
They probably show why if b^2 > 4ac there are two distinct real roots. However, you may just quote this in your actual exam.
7. I think it was looking for

x^2 + px + q = 0

(x + p/2)^2 - p^2/4 + q = 0

(x+p/2)^2 = p^2/4 - q

x + p/2 = +/- (p^2/4 - q)^1/2

x = -p/2 +/- (p^2/4 - q)^1/2

you can see that that will only be real when p^2/4 - q > 0

or p^2 > 4q

QED
8. (Original post by fishpaste)
I think it was looking for

x^2 + px + q = 0

(x + p/2)^2 - p^2/4 + q = 0

(x+p/2)^2 = p^2/4 - q

x + p/2 = +/- (p^2/4 - q)^1/2

x = -p/2 +/- (p^2/4 - q)^1/2

you can see that that will only be real when p^2/4 - q > 0

or p^2 > 4q

QED

what the hells this is this in the syllabus?
9. (Original post by TheWolf)
what the hells this is this in the syllabus?
Making x the subject and obtaining the quadratic formula, duuh.
10. (Original post by rachio)
is that all it is? the text book i have has some very very long complicated method which i do not understand. but if this is it it doesn't seem too hard.

thank you!
Well, the longer method is to complete the square

x^2 + px + q = 0
( x^2 + px + p^2/4 ) - p^2/4 + q = 0
(x + p/2)^2 = p^2/4 - q
x + p/2 = sqrt( p^2/4 - q )

Now we consider the square root. Square root does not have a real value if p^2/4 - q is negative. The square root has only one value if p^2/4 - q = 0. We require the equation to have distinct real roots, so that means that the square root must be real and have more than one value. This means that p^2/4 - q > 0, therefore p^2 > 4q.

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