What is the value of a ?

What is the value of a ?

1\(.a^2\) = \(b^2\)
\(|a|=|b|\) so or \(a=b\) or \(a=-b\), we have no value of b. Not sufficient

2.b - a = -12
Clearly not sufficient.

1+2
\(a-b=12\), so from statement 1 we can deduce that a=-b ( it cannot be a=b because a-b=b-b=0 and not 12).
\(-b-b=12,b=-6\) and \(a=6\). C

From F.S 1, we have (a-b)(a+b) = 0. Insufficient.

From F.S 2, we have b-a = -12. Insufficient.

Taking both together, we know that (a-b) is not equal to zero. Thus, (a+b) = 0, and we can get the value for a. Sufficient.

C.

Statement 1: \(a^2 = b^2\)
This implies, a=+b or -b
Insufficient.

Statement 2: b - a = -12
Clearly inssuficient.

Statement 1 and 2 Together:

Case 1: a=+b
b-b=-12
0=-12
Absurd Result.

Case 2: a=-b
b+b=-12
b=-6

Hence, C is our answer.

To prove that each statement alone is insufficient, we present a BIFURCATION (algebraic, in this case):

\(? = a\)

\(\left( 1 \right)\,\,{a^2} = {b^2}\)

\(Take\,\,\left\{ \begin{gathered}\\
{a^2} = {b^2} = 0\,\,\,\,\, \Rightarrow \,\,\,\,? = 0 \hfill \\\\
{a^2} = {b^2} = 1\,\,\,\,\, \Rightarrow \,\,\,\,? \ne 0\,\,\,\,\,\,\,\,\left( {a = \pm \,1} \right) \hfill \\ \\
\end{gathered} \right.\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{\text{INSUF}}{\text{.}}\)

\(\left( 2 \right)\,\,a\, - b = 12\)

\(Take\,\,\left\{ \begin{gathered}\\
b = 0\,\,\,\,\, \Rightarrow \,\,\,\,? = 12 \hfill \\\\
b = 1\,\,\,\, \Rightarrow \,\,\,\,? \ne 12\,\,\,\,\,\,\,\,\left( {a = \,13} \right) \hfill \\ \\
\end{gathered} \right.\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{\text{INSUF}}{\text{.}}\)

\(\left( {1 + 2} \right)\,\,\,\,\,\left\{ \begin{gathered}\\
{a^2} - {b^2} = 0\,\,\,\,\, \hfill \\\\
a - b = 12 \hfill \\ \\
\end{gathered} \right. \Rightarrow \,\,\,\,\,\,\left\{ \begin{gathered}\\
\left( {a + b} \right)\left( {a - b} \right) = 0 \hfill \\\\
a - b = 12 \hfill \\ \\
\end{gathered} \right.\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\{ \begin{gathered}\\
a + b = 0 \hfill \\\\
a - b = 12 \hfill \\ \\
\end{gathered} \right.\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( + \right)} \,\,\,\,\,2a = 12\,\,\,\,\,\, \Rightarrow \,\,\,\,\,a\,\,{\text{unique}} \Rightarrow \,\,\,\,\,\,{\text{SUF}}{\text{.}}\,\)


The above follows the notations and rationale taught in the GMATH method.

(1) \(a ^2 = b^2\)

\(a ^2 - b^2=0\)

\((a-b)(a+b)=0 \)
\(a-b=0\)
\(a=b\)

OR,
\(a+b=0\)
\(a=-b\)

(2) b − a = −12; Insufficient.

Considering both:

a=b, so, b-(-b)=12, 2b=12, b=6; Sufficient.

The answer is C