Question

if a and b are roots of the quadratic equation ax^2+bx+c=0, then quadratic equation ax^2-bx(x-1)+c(x-1)^2=0 has roots ?

Answer 1

@Abhisar ?
@bek_love ?
@Compassionate ?
@DollyAcquah ?
@emmigrace222 ?
@ganeshie8 ?

Answer 2

Well then

Answer 3

how to proceed
?

Answer 4

sorry the roots are

Answer 5

\[\alpha,\beta\]

Answer 6

honestly i dont know

Answer 7

i wrote \[ax^2+bx+c=a(x-\alpha)(x-\beta)\]

Answer 8

thats really good :)

Answer 9

then how to proceed?

Answer 10

ax^2-bx(x-1)+c(x-1)^2=0

Answer 11

should i expand it?

Answer 12

like it would be \[x^2(a-b+c)+x(b-2c)+c=0\]

Answer 13

or\[a(x-\alpha)(x-\beta)-bx^2+cx^2-2cx=0\]

Answer 14

@myininaya ?
@mathmale ?

Answer 15

@nincompoop ?

Answer 16

@oliviagutz ?

Answer 17

@Hotchellerae21 ?

Answer 18

@Future_UsArmy_MP ?

Answer 19

@Questionmachine ?

Answer 20

i am here

Answer 21

thanks can u suggest some way i have solved little bit bu how to proceed further?

Answer 22

ok what do u have so far

Answer 23

scroll upwards!

Answer 24

ok

Answer 25

maybe this site will help

Answer 26

http://www.rapidtables.com/math/algebra/Quadratic_equation.htm

Answer 27

@ganeshie8 ? can u suggest also please
thanks @Hotchellerae21 for ur effort

Answer 28

oh ok no problem @cody_123

Answer 29

@Abhisar ?
@beccamarx19 ?
@Compassionate ?
@dan815 ?
@Future_UsArmy_MP ?
@Hero ?

Answer 30

\(ax^2+bx+c=0\)

\(c/a = p \implies c = a\alpha\)
\(-b/a = q \implies b = -a\beta\)

Answer 31

\(\large ax^2-bx(x-1)+c(x-1)^2= x^2(a-b+c) + x(b-2c) + c \)

Answer 32

eliminate b, c by substituting their values ^

Answer 33

have you taken \[\alpha+\beta=p=-b/a\]?

Answer 34

@ganeshie8 ?

Answer 35

@ganeshie8 ?