Question

Solve (x-1)(x-2)=(k-1)(k-2), where k is a constant.

Answer 1

Try expanding the L.H.S and R.H.S

Answer 2

then...

Answer 3

tell me !

Answer 4

this is not the answer, the answer is x=k or x=3-k

Answer 5

what is the solution, please help

Answer 6

2x+2k-6=

Answer 7

x should be =k that is clearly visible just compare the terms more of the brains should be applied on the second part

Answer 8

@vernonwan: as ╰☆╮Openstudier╰☆╮ has already suggested: Try expanding the L.H.S and R.H.S: (x-1)(x-2) = (k-1)(k-2) becomes what? Note that both expansions will have 2 as their 3rd term. Would you do this work now, please?

Answer 9

\[x ^{2}-3x+2 = k ^{2} -3k +2\]

Answer 10

Now cancel the 2 from both sides, and subtract (k^2-3k) from both sides next.
You'll then have a quadratic equation in x. Try applying the quadratic formula to this quadratic equation.

Answer 11

As you said, x = 3-k is one of the solution
I replace to test
(3-k)^2 -3(3-k)+2 = 9 -6k+k^2-9 +6k+2 = k^2 +2
that is the left hand side . Question: do you really thing that k^2 +2 = k^2 +3k +2 ???? where k is a constant?

Answer 12

*think

Answer 13

I make that argument because your solution is not correct ( I mean the solution of x = 3-k). To me, the only one solution is x =k but you didn't accept it.

Answer 14

thanks, I am trying , the answer is from my text book

Answer 15

\[\frac{ x ^{2}-3x }{ k ^{2}-3k }=1\]

Answer 16

Excuse my insistence, but would you please follow the suggestions I gave you earlier. You can solve this problem using the quadratic formula. Obtaining the solution comes first. Checking the solutions comes second (but is an important step).

Answer 17

Your answer from the textbook is absolutely correct i worked it out

Answer 18

then,
\[x ^{2}-3x-k ^{2}+3k\]
how to solve this

Answer 19

I got it!

Answer 20

I got it!

Answer 21

x^2 -3x = k^2 - 3k
x^2 - k^2 = -3k + 3x
(x-k)(x+k)= 3 (-k+x)
now cancel out!!!
x+k=3
x=k-3


yesssssssssss

Answer 22

@mathmale i got it !

Answer 23

\[x ^{2}-3x-k ^{2}+3k=0\]has the form of a quadratic equation, right? A reminder: k is a constant.

You might want to re-write this as \[x ^{2}-3x+3k-k ^{2}=0\]
and then (referring to ax^2 + bx + c = 0), identify your
a as being equal to 1;
b as being equal to -3; and
c as being equal to 3k-k^2.

Now substitute these a, b and c into the quadratic formula. If done correctly, this will give you the two solutions x=k and x=3-k.

Answer 24

i had a simple method

Answer 25

but your method is also good

Answer 26

is my method correcct @mathmale

Answer 27

@vernonwan did u get my method??

Answer 28

x+k=3
x=k-3 is correct. What about the other solution, x=k?

To be correct, any method must result in ALL solutions.

Answer 29

i gave the method for the other sollution

Answer 30

already

Answer 31

yes thanks all of you