Problem of the Day


Let a,b,c, and d represent real numbers. Is ac+bd+bc+ad>0?

(1) ac^2+bd^2+bc^2+ad^2>0

(2) a^2d+b^2c+a^2c+b^2d<0

Reveal Answer

Answer

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:

$ac+bc+ad+bd$      Rearrange terms.

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

$(a+b)(c+d)$               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.


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