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M1ke
The stem gives us: is x=3 v x=-2?

Statement 1 gives us: if x=-3 v x=2 then the equation is not equal to zero. This statement adds no value to what we have to find out.

Statement 2 gives us: if x=-2 v x=-3 then the equation is not equal to zero. Because we had two possibilities in our stem: x=3 v x=-2, we can now cross out x=-2 because statement 2 tells us that the equation will not be equal to zero if x=-2.

The answer is B, with statement 2 alone we can state with certainty that x=3.


I can't get the explanation because my basic is not up to the mark. What should I do?
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If (x - 3)(x + 2) = 0, then what is the value of x?

Given that (x - 3)(x + 2) = 0. So, x = 3 or x = -2.

(1) (x + 3)(x - 2) ≠ 0. x ≠ -3 and x ≠ 2. This does not help to find whether x = 3 or x = -2. Not sufficient.


(2) (2x + 4)(2x + 6) ≠ 0. x ≠ -2 or x ≠ -3. Since x ≠ -2, then x must be equal to 3 (because from the stem we know that x = 3 or x = -2). Sufficient.

Answer: B.
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If (x-3)(x+2) = 0, then what is the value of x?

(1) (x+3)(x-2) ≠ 0

(2) (2x+4)(2x+6) ≠ 0

x=3 or x=-2

(1) (x+3)(x-2) ≠ 0
\( x \neq -3 and x \neq 2\)
NOT SUFFICIENT

(2) (2x+4)(2x+6) ≠ 0
\( x \neq -2 and x \neq -3\)
x = 3
SUFFICIENT

IMO B

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If (x-3)(x+2) = 0, then what is the value of x?

(1) (x+3)(x-2) ≠ 0

(2) (2x+4)(2x+6) ≠ 0


Can the problem above be solved by wavy-line method?
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Turjoy98
If (x-3)(x+2) = 0, then what is the value of x?

(1) (x+3)(x-2) ≠ 0

(2) (2x+4)(2x+6) ≠ 0


Can the problem above be solved by wavy-line method?

Yes, it can be solved, but it would take way more time as opposed to the direct method:

The equation gives us roots as 3, -2

A) X can't be equal to -3 and 2 (insuff.)

B) X can't be equal to -2 and -3, and from the initial equation we know x is either 3 or -2, and as x cant take -2, therefore x=3 (suff)

Point to be noted, the Wavy-line method is more useful for dealing with inequalities than direct roots.
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