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


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

See the Solution



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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