Question

If x,y are the roots of ax^2+bx+c=0 , find the equation with roots (x-y)^2 , (x+y)^2
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ANSWER= a^4x^2-2a^2(b^2-2ac)x+b^2(b^2-4ac)=0

Answer 1

find x^2+y^2
then (x-y)^2 = x^2-2xy+y^2
you know xy ...and x+y
so u know the roots.

Answer 2

(x+y)^2-2xy-2xy+(x+y)^2-2xy+2xy

Answer 3

is this rgight?

Answer 4

*right

Answer 5

what was that ?

Answer 6

I expanded all

Answer 7

x+y = -b/a
xy = c/a
x^2+y^2 =... ?

Answer 8

x^2-2xy+y^2

Answer 9

that is (x-y)^2 ...

Answer 10

but we did it last time a^2+b^2=(a+b)^2-2ab

Answer 11

ok...
x^2+y^2 = (x+y)^2 -xy = b^2/a^2 - c/a
got this ?

Answer 12

yup

Answer 13

-xy?

Answer 14

sorry, -2xy

Answer 15

ok ok

Answer 16

got it

Answer 17

b^2-2ac/a^2

Answer 18

(x-y)^2 = x^2-2xy+y^2 = b^2/a^2 -2c/a -2c/a
right ?
and (x+y)^2 = b^2/a^2
so you now have the 2 roots!

Answer 19

b^2-4ac/a^2

Answer 20

and whats that ?

Answer 21

mistake

Answer 22

why 2c/a 2 times

Answer 23

-2c/a-2c/a

Answer 24

(x-y)^2 = x^2-2xy+y^2 =[x^2+y^2]-2[xy]= [ b^2/a^2 -2c/a] -2[c/a]

Answer 25

(x+y)^2-2xy-2xy+(x+y)^2-2xy+2xy

Answer 26

i am not sure, whats that^
you always only write an expression (one side)
not equation

Answer 27

i expanded all (x-y)^2 + (x+y)^2

Answer 28

(x-y)^2 + (x+y)^2 = 2 (x^2+y^2)

Answer 29

I think its wrong

Answer 30

leave it... its wrong

Answer 31

so we were here => b^2/a^2-2c/a-2c/a

Answer 32

it forms b^2-2ac/a^2 am I correct?

Answer 33

(b^2-4ac)/a^2

Answer 34

yes exactly!! sorry

Answer 35

what about (x+y)^2

Answer 36

x+y= -b/a
(x+y)^2 = b^2/a^2

Answer 37

(b^2-4ac)/a^2 + b^2/a^2

Answer 38

the denominator should be a^4

Answer 39

for sum of roots, denominator = a^2 is correct.

Answer 40

ok

Answer 41

so the sum of roots is 2b^2-4ac / a^2

Answer 42

yes.

Answer 43

product of roots --> (x-y)^2 * (x+y)^2

Answer 44

how to do it?

Answer 45

you know both the roots...
b^2/a^2 and (b^2-4ac)/a^2
just take this product...

Answer 46

yes

Answer 47

product of roots = b^2(b^2-4ac)/a^4
now can you get to the answer ?

Answer 48

yes

Answer 49

ok, good :)