Question

Question for quadratics

Answer 1

express that equation as ax^2+bx+c =0
use b^2-4ac =0

Answer 2

How can I even make this in the standard form?

Answer 3

expand left side and right side.

Answer 4

Can i cross out the (x-1) on both sides?

Answer 5

yes, sure.

Answer 6

how how can I set it in standard from?

Answer 7

let me try without cancelling...

Answer 8

i am getting imaginary values of k :O

Answer 9

@ganeshie8 any thoughts?

Answer 10

thinking... @phi

Answer 11

we wanto prove , D = 0 ?

Answer 12

x=1 satisfies the equation, but the value of k....... ?

Answer 13

i guess we need to prove D=0 for only one value of k

Answer 14

it seems tricky lol..

Answer 15

let me try again,
kx^2-k = x^2-2x+1
(k-1)x^2 +2x + (-k-1) = 0
oh, i got my error.

Answer 16

i got d same
wats wrong wid that hnmm

Answer 17

damn
4 = -4 (k-1) (k+1)
-1 = k^2-1
k^2=0
yup, the only value of k is 0

Answer 18

damn yes k=0 is the oly value of k ugh

Answer 19

let me write all steps together.,
for that equation to have only one solution, D = 0

kx^2-k = x^2-2x+1
(k-1)x^2 +2x + (-k-1) = 0
a=k-1, b=2 , c= -k-1

b^2=4ac
4 = 4 (k-1) (-k-1)
4 = -4 (k-1) (k+1)
-1 = k^2-1
k^2=0
k= 0 is the only value of k.

Answer 20

thank you for your help guys

Answer 21

(k-1)x^2 +2x + (-k-1) = 0

How did you get that @hartnn

Answer 22

kx^2-k = x^2-2x+1
kx^2 -x^2 +2x -k-1 = 0

subtracting x^2-2x+1 from both sides...