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quadratic equation (1 Viewer)

sungkwo

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consider the quadratic equation: ax^2+px+aq+q=0
where a does not equal zero& 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
 

gurmies

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I believe here you've made an error in the question:

ax^2+px+aq+q=0 is the equation.

You have said that x = 1 solves this equation:

a + p + aq + q = 0

p + q = -a - aq

p + q = -a(1+q)

Therefore, unless you somehow know q = -1, p + q =/ 0, as a =/ 0.
 

jockrussell

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where did u get this question from.
I tried playing with it but after 4 pages i got lost lol.
I tried using the quadratic formula to try find something and subbing 1 into the original equation and then tried some manipulation of algebraic expressions.
 

jockrussell

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if it helps any1 else i found that -1 is the other root of the equation

and i found p to be 0

and if im right that means q = 0, just taking the answer p+q = 0
And if q=0 then the question is wrong because p+q=-a(q+1) where a cannot equal zero will end up not equaling zero. But im probably wrong
 
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