How do you solve #sqrt(x+3)-sqrt(x-1)=1#?

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Answer 1 · 2015-04-19T10:40:53

the answer is #x=13/4#

First, let's call #a=sqrt(x+3)# and #b=sqrt(x-1)#

Now use the rule #(a+b)(a-b)=a^2-b^2#
so #4=a^2-b^2=a+b#

So we have #a+b=4# and #a-b=1#

We sum and subtract respectively the two equations and we have

#2a=5#
#2b=3#

So #sqrt(x+3)=5/2 => x = +-25/4 - 3 =>x \in {13/4,-37/4}#
And #sqrt(x-1)=3/2 => x= +- 9/4 + 1 => x \in {13/4,-5/4}#

so #x=13/4#

Answer 2 · 2015-04-19T13:57:03

Firstly know that:

#sqrt(a)sqrt(b)=sqrt(ab)#

And also that:

#sqrt(q)sqrt(q)=q#

Knowing this, let's find the value of x...

#sqrt(x+3)-sqrt(x-1)=1#

#(sqrt(x+3)-sqrt(x-1))^2=1^2#

#(sqrt(x+3)-sqrt(x-1))(sqrt(x+3)-sqrt(x-1))=1#

#x+3-sqrt(x+3)sqrt(x-1)-sqrt(x+3)sqrt(x-1)+(x-1)=1#

#x+3-2sqrt(x+3)sqrt(x-1)+x-1=1#

#2x+2-2sqrt(x+3)sqrt(x-1)=1#

#2(x+1-sqrt(x+3)sqrt(x-1))=1#

#x+1-sqrt(x+3)sqrt(x-1)=1/2#

#x+1-1/2=sqrt((x+3)(x-1))#

#x+1/2=sqrt((x+3)(x-1))#

#(x+1/2)^2=(x+3)(x-1)#

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

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

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

#:. x=3+1/4#