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If x(x + 1) = x + 1, what are all the possible values of x ?

A. -1 only
B. 0 only
C. 1 only
D. -1 and 0
E. -1 and 1 --> correct: x(x + 1) = x + 1 => x^2+x=x+1 =>x^2=1 => x=+1 or -1
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Bunuel
If x(x + 1) = x + 1, what are all the possible values of x ?

A. -1 only
B. 0 only
C. 1 only
D. -1 and 0
E. -1 and 1


PS21184

RULE: Don't cancel the variable part (x+1 in this case)unless we know that it's non-zero

x(x + 1) - (x + 1) = (x+1)*(x-1) = 0

i.e. either x+1 = 0 or x-1 = 0

i.e. either x = -1 or x = 1

ANswer: Option E
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Bunuel
If \(x(x + 1) = x + 1\), what are all the possible values of x ?

A. -1 only
B. 0 only
C. 1 only
D. -1 and 0
E. -1 and 1


PS21184
\(x(x + 1) = x + 1\)

Or, \(x^2 + x = x + 1\)

Or, \(x^2 = 1\)

Hence, \(x = -1\) & \(x = +1\), Answer must be (E)
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Bunuel
If x(x + 1) = x + 1, what are all the possible values of x ?

A. -1 only
B. 0 only
C. 1 only
D. -1 and 0
E. -1 and 1
PS21184


Given: x(x + 1) = x + 1
Note: Some students will be tempted to divide both sides of the equation by (x + 1) to get x = 1.
However, since we don't know the value of x, we may be inadvertently dividing both sides by 0, which is a big no-no

Instead, let's first expand the left side to get: x² + x = x + 1
Subtract x from both sides: x² = 1
From here we can see that, EITHER x = 1 OR x = -1

Answer: E

Cheers,
Brent
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x(x + 1) = x + 1

Do not cancel the (x+1) from both sides, its a trap.

Subtract (x+1) from both sides

x(x+1) - (x+1) = (x+1) - (x+1)
(x-1)(x+1) = 0

==> x = +1, -1

E.
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Bunuel
If x(x + 1) = x + 1, what are all the possible values of x ?

A. -1 only
B. 0 only
C. 1 only
D. -1 and 0
E. -1 and 1


If x = -1, we see that both sides of the equation will be 0. Thus, a value of x is -1.

If x ≠ -1, divide both sides of the equation by x + 1 and we’ll have x = 1. Thus, another value of x is 1.

Answer: E
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\(x(x + 1) = x + 1\)
\(x(x + 1) - (x + 1) = 0\)
\((x+1)(x-1)=0\)
\(x=-1\) or \(x=1\)

Answer E
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Bunuel
If x(x + 1) = x + 1, what are all the possible values of x ?

A. -1 only
B. 0 only
C. 1 only
D. -1 and 0
E. -1 and 1


PS21184

Asked: If x(x + 1) = x + 1, what are all the possible values of x ?

x(x + 1) = x + 1
(x+1)(x-1) = 0

x = 1 or -1

IMO E
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