Solve the simultaneous equations y = √#x#+2 and (#y+x#)(#y-x#)=0 ?

1 Answer
Nov 18, 2017

#(-1,1),(2,2)#

Explanation:

#y=sqrt(x+2)to(1)#

#(y+x)(y-x)=0larrcolor(blue)"factors of difference of squares"#

#rArry^2-x^2=0to(2)#

#color(blue)"substitute "y=sqrt(x+2)" into equation "(2)#

#(sqrt(x+2))^2-x^2=0#

#>rArrx+2-x^2=0#

#"multiply through by "-1#

#x^2-x-2=0larrcolor(blue)"in standard form"#

#"the factors of - 2 which sum to - 1 are +1 and - 2"#

#rArr(x+1)(x-2)=0#

#"equate each factor to zero and solve for x"#

#x+1=0rArrx=-1#

#x-2=0rArrx=2#

#"substitute these values into equation "(1)#

#x=-1toy=sqrt(-1+2)=1#

#x=2toy=sqrt(2+2)=2#

#"points of intersection are "(-1,1)" and "(2,2)#