If bc ≠ 0, what is the value of (a^2 - b^2 - c^2)/bc?

Bunuel:
According to second option: The value of expression will always be always be zero.
Right??

Hi Shree9975,

You have a theory, so you should try to prove that it's true. Try TESTing VALUES....

Choose values for A, B and C that "fit" the equation in Fact 2:

A+B+C = 0

Then plug those values into the question....Then choose a different set of values to TEST and plug those in....You don't need Bunuel to figure out if your theory is correct.

GMAT assassins aren't born, they're made,
Rich

If bc ? 0, what is the value of \(\frac{a^2 - b^2 - c^2}{bc}\)?

Stmt (1) |a| = 1, |b| = 2, |c| = 3

note: a^2, b^2, c^2 will be {+ve} , while b * c can be {+ve or -ve} 6

\(= \frac{1 - 4 - 9}{\pm 6}\\
= \pm 2\)

insufficient

Stmt (2) a + b + c = 0

we know bc is in denominator ; lets write a in terms of (b+c)

\(a = -(b+c)\) --- let's square both sides

\(a^2 = b^2 + c^2 + 2bc\) --- plug in this value of a in main equation

\(\frac{a^2 - b^2 - c^2}{bc} \\
= \frac{b^2 + 2bc + c^2 - b^2 - c^2}{bc} \\
= \frac{2bc}{bc} = 2\)

sufficient

Ans: B

MANHATTAN GMAT OFFICIAL SOLUTION:

(1) INSUFFICIENT: The absolute value signs tell us that a = ±1, b = ±2, and c = ±3. In the numerator, each variable is squared, so their signs are irrelevant:

\(a^2 - b^2 - c^2 = 1^2 - 2^2 - 3^2 = 1 -4 - 9 = -12\)

However, while |b||c| = (2)(3) = 6, the denominator bc could be either 6 or –6, depending on the signs of b and c. Therefore:
\(\frac{a^2 - b^2 - c^2}{bc}=2\) OR -2.

(2) SUFFICIENT: Given that a + b + c = 0, we know that a = - (b + c). Substitute this value of a into the expression in the question, and simplify, using one of the quadratic special products along the way:

\(\frac{a^2 - b^2 - c^2}{bc}=\frac{(-(b+c))^2 - b^2 - c^2}{bc}=\)
\(=\frac{(b+c)^2 - b^2 - c^2}{bc}=\)
\(=\frac{b^2+2bc+c^2 - b^2 - c^2}{bc}=\)
\(=\frac{2bc}{bc}=2\)

The correct answer is B.

(a-b-c)(a+b+c)+2bc / bc = 2

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