having trouble on it
I assume it has parentheses like this:
(x+1)=√(x+3)
square both sides
(x+1)^2 = x+3
expand
x^2 + 2x + 1 = x + 3
make it a quadratic equation
x^2 + x - 2 = 0
solve
(x+2)(x-1) = 0
x = -2 or x = 1
plug both back in and they BOTH work (ignore answer above mine, and remember that √(-2 + 3) = ±1)
(x+1)^2=[âx+3]^2
x^2+2x+1=x+3
x^2+x-2=0
(x-1)(x+2)=0
x=1,-2
x=1
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Verified answer
I assume it has parentheses like this:
(x+1)=√(x+3)
square both sides
(x+1)^2 = x+3
expand
x^2 + 2x + 1 = x + 3
make it a quadratic equation
x^2 + x - 2 = 0
solve
(x+2)(x-1) = 0
x = -2 or x = 1
plug both back in and they BOTH work (ignore answer above mine, and remember that √(-2 + 3) = ±1)
(x+1)^2=[âx+3]^2
x^2+2x+1=x+3
x^2+x-2=0
(x-1)(x+2)=0
x=1,-2
x=1