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I need help with my algebra 2. I also want to know how to solve this.

I need help with my algebra 2. I also want to know how to solve this.-example-1
User Alea
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1 Answer

4 votes

Add 1 to both sides:


√(x+3) = x+1

In cases like this, we have to remember that a root is always positive, so we can square both sides only assuming that


x+1 \geq 0 \iff x \geq -1

Under this assumption, we square both sides and we have


x+3 = (x+1)^2 \iff x+3 = x^2+2x+1 \iff x^2+x-2 = 0

The solutions to this equation are


x = -2,\ x=1

But since we can only accept solutions greater than -1, we discard
x=-2 and accept
x=1.

In fact, we have


x=-2 \implies √(-2+3)-1=0\\eq -2

and


x=1 \implies √(4)-1=1

which is the only solution.

User Frunkad
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