How do you simplify #(sqrtx-y)(sqrtx+y)#?

2 Answers
May 1, 2017

Answer:

See the solution process below:

Explanation:

This problem is a special form and can follow the rule:

#(a - b)(a + b) = a^2 - b^2#

Substituting #sqrt(x)# for #a# and #y# for #b# gives:

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

We can also multiply these two terms the long way to obtain the same answer. To multiply these two terms you multiply each individual term in the left parenthesis by each individual term in the right parenthesis.

#(color(red)(sqrt(x)) - color(red)(y))(color(blue)(sqrt(x)) + color(blue)(y))# becomes:

#(color(red)(sqrt(x)) xx color(blue)(sqrt(x))) + (color(red)(sqrt(x)) xx color(blue)(y)) - (color(red)(y) xx color(blue)(sqrt(x))) - (color(red)(y) xx color(blue)(y))#

#(sqrt(x))^2 + ysqrt(x) - ysqrt(x) - y^2#

#x + 0 - y^2#

#x - y^2#

May 1, 2017

Answer:

#(sqrtx-y)(sqrtx+y)=color(red)(x-y^2)#

Explanation:

In general
#color(white)("XXX")(a-b)(a+b)=a^2-b^2#

If we let #a=sqrt(x)#
then #a^2=x#

and if we let #b=y#
then #b^2=y^2#

So #(a-b)(a+b)=a^2-b^2#
is equivalent to #(sqrtx-y)(sqrtx+y)=x-y^2#