# Problem of the Day

Let and represent real numbers. Is ?

(1) (2) C. Statements (1) and (2) together are sufficient.

See the Solution

### Solution

[latexpage]

The key to this problem is realizing that all of these expressions can be factored by grouping. We want to know if \$ac+bd+bc+ad>0\$. Factor the left side of the inequality as follows:

\$ac+bc+ad+bd\$ Â  Â  Â Rearrange terms.

\$c(a+b)+d(a+b)\$ Â  Â Factor out a \$c\$ and a \$d\$.

\$(a+b)(c+d)\$ Â  Â  Â  Â  Â  Â  Â  Factor out \$(a+b)\$ from both terms.

Now we are trying to determine if \$(a+b)(c+d)>0\$. Let’s look at the statements:

Statement 1: INSUFFICIENT. Statement 1 can be factored in a similar fashion to produce the equivalent expression \$(a+b)(c^2+d^2)>0\$. \$c^2+d^2\$ must be positive because it is the sum of two squares. We know it is not equal to zero because otherwise this inequality would be false. Given that \$c^2+d^2\$ is positive, \$(a+b)\$ must also be positive. However, we still have no information about the factor \$(c+d)\$, so we cannot solve the problem yet.

Statement 2: INSUFFICIENT. Statement 2 can also be factored in a similar fashion to produce the expression \$(c+d)(a^2+b^2)<0\$. Similar reasoning to that employed above shows that this is only true if \$(c+d)\$ is negative. However, we know nothing about \$(a+b)\$, so the statement is insufficient.

Statements 1&2: SUFFICIENT. Combining the fact that \$(a+b)\$ is positive and \$(c+d)\$ is negative, tells us that \$(a+b)(c+d)\$ can only be negative. Therefore, both statements together are sufficient.