anonymous
  • anonymous
Given that one root of the equation x^2 + 2px - q = o is twice the other, express q in terms of p.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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amistre64
  • amistre64
you recall the quadratic formula?
anonymous
  • anonymous
We were told not to do that method as it is extremely long and time consuming
anonymous
  • anonymous
Can you show me the sum of roots and product of roots method of doing it?

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.Sam.
  • .Sam.
can we use teh sum and product of roots?
anonymous
  • anonymous
i think so...
amistre64
  • amistre64
im do not recall hearing of those kinds of methods. i do not see how the quad formula is long and consuming tho .... \[\frac{-b-\sqrt{b^2-4ac}}{a}=\frac{-b+\sqrt{b^2-4ac}}{2a}\]
anonymous
  • anonymous
Alright, sure thing i will try to understand the steps.
amistre64
  • amistre64
in any case, the roots will still be a result of the quad formula .... not too sure how you would be able to define them otherwise
anonymous
  • anonymous
Can you show me the steps please?
amistre64
  • amistre64
\[\frac{-b-\sqrt{b^2-4ac}}{a}=\frac{-b+\sqrt{b^2-4ac}}{2a}\] \[-2b-2\sqrt{b^2-4ac}=-b+\sqrt{b^2-4ac}\] \[-2(2p)-2\sqrt{(2p)^2+4q}=-(2p)+\sqrt{(2p)^2+4q}\] \[-2p=3\sqrt{(2p)^2+4q}\] \[-\frac23p=\sqrt{4p^2+4q}\] \[-\frac23p=2\sqrt{p^2+q}\] \[-\frac13p=\sqrt{p^2+q}\] \[\frac19p^2=p^2+q\] \[-\frac89p^2=q\] maybe
anonymous
  • anonymous
I see so this is how i do it. Thank you :D *Thumbs up*
anonymous
  • anonymous
Thanks dude.
amistre64
  • amistre64
this is how i would approach it, yes.
anonymous
  • anonymous
Can i ask you to help me solve a similar question?
amistre64
  • amistre64
dunno, i only have so many good ideas in a day ...
anonymous
  • anonymous
in 3 mins time
anonymous
  • anonymous
Can i swop the denominator ?
anonymous
  • anonymous
Will it affect the end result?
amistre64
  • amistre64
what do you mean be "swap the denominator"?
anonymous
  • anonymous
the a and 2a swop the position
anonymous
  • anonymous
because gen formula is -b positive/negative square root (b square -4ac) over 2a
anonymous
  • anonymous
The denominator represent the roots and a is the unknown variable right?
amistre64
  • amistre64
the quadratic formula has 2 parts to it, a vertex axis called a discriminant, and a +- part that measures an equal distance to both sides of that axis to define the zeros: r1 <= vertex <= r2 \[\frac{-b}{2a}-\frac{\sqrt{b^2-4ac}}{2a}~\le~\frac{-b}{2a}~\le~\frac{-b}{2a}+\frac{\sqrt{b^2-4ac}}{2a}\]
amistre64
  • amistre64
the phrase "2 times as large" suggests to me that the left side has to double to equal the right side
anonymous
  • anonymous
what if it's like the root differ by 2
anonymous
  • anonymous
or you need to see the whole question
amistre64
  • amistre64
the left and right sides of the middle are the roots; if we keep it clean and just call them r1 and r2 |r1 - r2| = 2 expresses the difference between them as "2"
anonymous
  • anonymous
I see wow this logic is quite thought provoking..
anonymous
  • anonymous
okay moving on to the next question now

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