some random question | Revision Server | Page 1

hearty crow Jan 10, 2024, 11:33 PM

#

Pls help me I ended up with 6/7 - 9 = 1+ k/c however this is wrong

sleek basin Jan 10, 2024, 11:38 PM

#

The point is you are meant to equate coefficients not solve like a normal equation.

#

#

Like this

#

so you compare the two terms with the root and the terms without it

#

so 3root(c) = kroot(c) so k = 3

#

and then the terms without the roots are 2c - 9 and 1 so 2c - 9 = 1

#

so c = 5

#

which both are prime so it satisfies the question

#

it should show an identity symbol

hearty crow Jan 10, 2024, 11:41 PM

#

I see how I went wrong when I did root c times 2 root c I made it 3 root c and I didn’t realise 2c -9 can = 1 I thought it had to be 1

sleek basin Jan 10, 2024, 11:41 PM

#

ah

hearty crow Jan 10, 2024, 11:42 PM

#

How did u find out c = 5

#

I see

sleek basin Jan 10, 2024, 11:42 PM

#

2c - 9 = 1

hearty crow Jan 10, 2024, 11:42 PM

#

Ok thankyou

sleek basin Jan 10, 2024, 11:42 PM

#

add 9 and divide by 2

#some random question