groovy261 wrote:

What is the value of x?

1. 2x + 1 = 0

2. (x+1)^2 = x^2

The OG says that the second equation can be solved by expanding the squares and reducing it to 2x +1 = 0 thereby making it sufficient to solve. My question is why can't we solve it like this?

(x+1)^2 = x^2

Square root both sides and you get

(x+1) = x

subtract x from both sides to get

1 = 0

The equation now is unsolvable and therefore I guess insufficient.

Thanks

You cannot simply take square roots on both sides, because

\sqrt{x^2} is not necessarily equal to

x. Indeed, as you've shown, if they were always equal the equation could not possibly have any solutions, but it does: x = -1/2. In general:

\begin{align*}

\sqrt{x^2} &= x \text{ if } x \geq 0 \\

\sqrt{x^2} &= -x \text{ if } x < 0

\end{align*}You might try this with an example negative number. Taking -3, for example, notice that

\sqrt{(-3)^2} is not equal to -3; it's equal to

\sqrt{9} = 3, which is the same thing as

-(-3).

Knowing this, it becomes possible to answer the question by taking square roots on both sides, though the approach taken in the OG (expanding both sides) would be more straightforward. Still, if we want to take roots on both sides:

Just glancing at the equation in Statement 2, it may be clear that x and x+1 cannot both be positive; the equation would make no sense, since

(x+1)^2 would certainly be larger than

x. Similarly, x and x+1 cannot both be negative. So one of them must be negative, the other positive, and since x is smaller than x+1, x must be negative and x+1 must be positive. Since

x must be negative,

\sqrt{x^2} = -x. Now we can take square roots, knowing what sign we will need for each:

\begin{align*}

(x+1)^2 &= x^2 \\

\sqrt{(x+1)^2} &= \sqrt{x^2} \\

x+1 &= -x \\

x &= \frac{-1}{2}

\end{align*}
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