How do you solve #(2x+1)^2=(x+2)^2#?

1 Answer
May 19, 2017

Answer:

#x=+-1#

Explanation:

#"expand factors on both sides using the FOIL method"#

#4x^2+4x+1=x^2+4x+4#

#"subtract " x^2" from both sides"#

#4x^2-x^2+4x+1=cancel(x^2)cancel(-x^2)+4x+4#

#rArr3x^2+4x+1=4x+4#

#"subtracting 4x from both sides, gives"#

#3x^2+1=4#

#"subtract 4 from both sides"#

#3x^2+1-4=4-4#

#rArr3x^2-3=0larrcolor(blue)" quadratic equation"#

#3(x^2-1)=0larrcolor(blue)" common factor of 3"#

#3(x-1)(x+1)=0larrcolor(blue)" difference of aquares"#

#"equate each factor to zero and solve"#

#x-1=0rArrx=1#

#x+1=0rArrx=-1#

#color(blue)"As a check"#

Substitute these values into the equation and if both sides are equal then they are the solutions.

#x=1to(2+1)^2=(1+2)^2to" True"#

#x=-1to(-1)^2=(1)^2to" True"#

#rArrx=+-1" are the solutions"#