How do you solve sqrt(x+3)-sqrt(x-1)=1x+3x1=1?

2 Answers
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=x1

Now use the rule (a+b)(a-b)=a^2-b^2(a+b)(ab)=a2b2
so 4=a^2-b^2=a+b4=a2b2=a+b

So we have a+b=4a+b=4 and a-b=1ab=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=52x=±2543x{134,374}
And sqrt(x-1)=3/2 => x= +- 9/4 + 1 => x \in {13/4,-5/4}x1=32x=±94+1x{134,54}

so x=13/4x=134

Apr 19, 2015

Firstly know that:

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

And also that:

sqrt(q)sqrt(q)=qqq=q

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

sqrt(x+3)-sqrt(x-1)=1x+3x1=1

(sqrt(x+3)-sqrt(x-1))^2=1^2(x+3x1)2=12

(sqrt(x+3)-sqrt(x-1))(sqrt(x+3)-sqrt(x-1))=1(x+3x1)(x+3x1)=1

x+3-sqrt(x+3)sqrt(x-1)-sqrt(x+3)sqrt(x-1)+(x-1)=1x+3x+3x1x+3x1+(x1)=1

x+3-2sqrt(x+3)sqrt(x-1)+x-1=1x+32x+3x1+x1=1

2x+2-2sqrt(x+3)sqrt(x-1)=12x+22x+3x1=1

2(x+1-sqrt(x+3)sqrt(x-1))=12(x+1x+3x1)=1

x+1-sqrt(x+3)sqrt(x-1)=1/2x+1x+3x1=12

x+1-1/2=sqrt((x+3)(x-1))x+112=(x+3)(x1)

x+1/2=sqrt((x+3)(x-1))x+12=(x+3)(x1)

(x+1/2)^2=(x+3)(x-1)(x+12)2=(x+3)(x1)

x^2+1/2x+1/2x+1/4=x^2-x+3x-3x2+12x+12x+14=x2x+3x3

x^2+x+1/4=x^2+2x-3x2+x+14=x2+2x3

x^2-x^2+1/4+3=2x-xx2x2+14+3=2xx

:. x=3+1/4