Question

Given that one root of the equation x^2 + 2px - q = o is twice the other, express q in terms of p.

Answer 1

you recall the quadratic formula?

Answer 2

We were told not to do that method as it is extremely long and time consuming

Answer 3

Can you show me the sum of roots and product of roots method of doing it?

Answer 4

can we use teh sum and product of roots?

Answer 5

i think so...

Answer 6

im do not recall hearing of those kinds of methods. i do not see how the quad formula is long and consuming tho ....

\[\frac{-b-\sqrt{b^2-4ac}}{a}=\frac{-b+\sqrt{b^2-4ac}}{2a}\]

Answer 7

Alright, sure thing i will try to understand the steps.

Answer 8

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

Answer 9

Can you show me the steps please?

Answer 10

\[\frac{-b-\sqrt{b^2-4ac}}{a}=\frac{-b+\sqrt{b^2-4ac}}{2a}\]
\[-2b-2\sqrt{b^2-4ac}=-b+\sqrt{b^2-4ac}\]
\[-2(2p)-2\sqrt{(2p)^2+4q}=-(2p)+\sqrt{(2p)^2+4q}\]
\[-2p=3\sqrt{(2p)^2+4q}\]
\[-\frac23p=\sqrt{4p^2+4q}\]
\[-\frac23p=2\sqrt{p^2+q}\]
\[-\frac13p=\sqrt{p^2+q}\]
\[\frac19p^2=p^2+q\]
\[-\frac89p^2=q\]
maybe

Answer 11

I see so this is how i do it. Thank you :D *Thumbs up*

Answer 12

Thanks dude.

Answer 13

this is how i would approach it, yes.

Answer 14

Can i ask you to help me solve a similar question?

Answer 15

dunno, i only have so many good ideas in a day ...

Answer 16

in 3 mins time

Answer 17

Can i swop the denominator ?

Answer 18

Will it affect the end result?

Answer 19

what do you mean be "swap the denominator"?

Answer 20

the a and 2a swop the position

Answer 21

because gen formula is -b positive/negative square root (b square -4ac) over 2a

Answer 22

The denominator represent the roots and a is the unknown variable right?

Answer 23

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

\[\frac{-b}{2a}-\frac{\sqrt{b^2-4ac}}{2a}~\le~\frac{-b}{2a}~\le~\frac{-b}{2a}+\frac{\sqrt{b^2-4ac}}{2a}\]

Answer 24

the phrase "2 times as large" suggests to me that the left side has to double to equal the right side

Answer 25

what if it's like the root differ by 2

Answer 26

or you need to see the whole question

Answer 27

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"

Answer 28

I see wow this logic is quite thought provoking..

Answer 29

okay moving on to the next question now