The answer is: " x = 0, 1 " .
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Explanation:
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Given:
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" √(x + 1) − 1 = x " ; Solve for "x" ;
First, let us assume that "x ≥ -1 "
Add "1" to EACH SIDE of the equation:
→ √(x + 1) − 1 + 1 = x + 1 ;
to get:
→ √(x + 1) = x + 1 .
Now, "square" EACH side of the equation:
→ [√(x + 1) ]² = (x + 1 )² ;
to get:
x + 1 = (x + 1)²
↔ (x + 1)² = (x + 1) .
Expand the "left-hand side" of the equation:
→ (x + 1)² = (x + 1)(x +1) ;
Note: (a+b)(c+d) = ac +ad + bc + bd ;
As such: (x + 1)(x + 1) = (x*x) + (x*1) +(1(x) + (1*1) ;
= x² + 1x + 1x + 1 ;
= x² + 2x + 1 ;
Now, substitute this "expanded" value, and bring down the "right-hand side" of the equation; and rewrite the equation:
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" (x + 1)² = (x + 1)(x +1) " ;
→ Rewrite as: " x² + 2x + 1 = x + 1 " ;
Subtract "x" ; and subtract "1" ; from EACH SIDE of the equation:
→ x² + 2x + 1 - x - 1 = x + 1 - x - 1 ;
to get: → x² − x = 0
Factor out an "x" on the "left-hand side" of the equation:
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x² − x = x(x − 1) ;
→ x (x − 1) = 0 ;
We have: "x" and "(x − 1)" ; when either of these two multiplicands are equal to zero, then the "right-hand side of the equation equals "zero" .
So, one value of "x" is "0" .
The other value for "x" ;
→ x − 1 = 0 ;
Add "1" to each side of the equation:
→ x − 1 + 1 = 0 + 1 ;
→ x = 1 ;
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So, the answers:
" x = 0, 1 " .
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