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

Question

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

(1) |a| = 1, |b| = 2, |c| = 3
(2) a + b + c = 0

IanStewart · Accepted Answer

S1 cannot be sufficient, because we can make the denominator positive or negative without changing the value of the numerator.

For S2, if a+b+c = 0, then a = -b-c. So

\begin{align}
\frac{a^2 - b^2 - c^2}{bc} &= \frac{(-b-c)^2 - b^2 - c^2}{bc} \
&= \frac{(-1)^2(b + c)^2 - b^2 - c^2}{bc} \
&= \frac{b^2 + 2bc + c^2 - b^2 - c^2}{bc} \
&= \frac{2bc}{bc} \
&= 2
\end{align}

Plugging in values seems needlessly time-consuming here.

peachfuzz · Answer

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

The numerator of the expression will always equal -12

(1) |a| = 1, |b| = 2, |c| = 3
bc can equal 6 or -6.
The expression can equal 2 or -2
insufficient

(2) a + b + c = 0
Plugging in many different values and neg values the
Expression always equals 2
sufficient

Answer: B

Shree9975 · Answer

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

EMPOWERgmatRichC · Answer

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

UJs · Answer

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

Bunuel · Answer

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.

Sebastian Shoaib · Answer

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

bumpbot · Answer

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

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

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