How do you solve #-( y - 1) = ( x + 5) ^ { 2}#?

1 Answer
Kushagra
Jan 16, 2018

#-(y-1)" when opened"=-y+1#
A law:
#(a+b)^2=a^2+2ab+b^2#
#(x+5)^2=x^2+2xx x xx5+5^2#
#(x+5)^2=x^2+10x+25#
Apply it
#-y+1=x^2+10x+25#
Subtract 1 from both sides
#-y+cancel1-cancel1=x^2+10x+25-1#
#-y=x^2+10x+24#
You can leave it from here.
Factorize #x# only
#-y=x(x+10)+24#
Divide #x# from both sides
#-y/x=x+10+24/x#
It will go on and on

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