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

2 Answers
Alberto M.
Apr 19, 2015

the answer is x=13/4x=134

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

Now use the rule (a+b)(a-b)=a^2-b^2(a+b)(a−b)=a2−b2
so 4=a^2-b^2=a+b4=a2−b2=a+b

So we have a+b=4a+b=4 and a-b=1a−b=1

We sum and subtract respectively the two equations and we have

2a=52a=5
2b=32b=3

So sqrt(x+3)=5/2 => x = +-25/4 - 3 =>x \in {13/4,-37/4}√x+3=52⇒x=±254−3⇒x∈{134,−374}
And sqrt(x-1)=3/2 => x= +- 9/4 + 1 => x \in {13/4,-5/4}√x−1=32⇒x=±94+1⇒x∈{134,−54}

so x=13/4x=134

Tiago Hands
Apr 19, 2015

Firstly know that:

sqrt(a)sqrt(b)=sqrt(ab)√a√b=√ab

And also that:

sqrt(q)sqrt(q)=q√q√q=q

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

sqrt(x+3)-sqrt(x-1)=1√x+3−√x−1=1

(sqrt(x+3)-sqrt(x-1))^2=1^2(√x+3−√x−1)2=12

(sqrt(x+3)-sqrt(x-1))(sqrt(x+3)-sqrt(x-1))=1(√x+3−√x−1)(√x+3−√x−1)=1

x+3-sqrt(x+3)sqrt(x-1)-sqrt(x+3)sqrt(x-1)+(x-1)=1x+3−√x+3√x−1−√x+3√x−1+(x−1)=1

x+3-2sqrt(x+3)sqrt(x-1)+x-1=1x+3−2√x+3√x−1+x−1=1

2x+2-2sqrt(x+3)sqrt(x-1)=12x+2−2√x+3√x−1=1

2(x+1-sqrt(x+3)sqrt(x-1))=12(x+1−√x+3√x−1)=1

x+1-sqrt(x+3)sqrt(x-1)=1/2x+1−√x+3√x−1=12

x+1-1/2=sqrt((x+3)(x-1))x+1−12=√(x+3)(x−1)

x+1/2=sqrt((x+3)(x-1))x+12=√(x+3)(x−1)

(x+1/2)^2=(x+3)(x-1)(x+12)2=(x+3)(x−1)

x^2+1/2x+1/2x+1/4=x^2-x+3x-3x2+12x+12x+14=x2−x+3x−3

x^2+x+1/4=x^2+2x-3x2+x+14=x2+2x−3

x^2-x^2+1/4+3=2x-xx2−x2+14+3=2x−x

:. x=3+1/4

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