I have a quadratic expression, which I have factored to correctly be: $(x-9)(x-2) > 0$

However, I don't know how to determine the two values of X after this, the correct answers are x < 2, x > 9, however, is this because x is smaller, and 9 is larger, or is their a specific algebraic step in solving this?

  • $\begingroup$ For two numbers to give positive product, either both numbers are positive, or both numbers are negative. $\endgroup$ – peterwhy May 31 '14 at 13:38
  • $\begingroup$ That was the formal way of writing solution; I usually think faster in the following way. If you are familiar enough with quadratic curves, you know $y=(x-9)(x-2)$ gives a U-shaped curve. Knowing $x=2$ and $9$ are roots, which part can then give $y>0$? Both ends, $x<2$ or $x>9$, instead of $2<x<9$. $\endgroup$ – peterwhy May 31 '14 at 13:46

On the left-hand side you have a product of two numbers. For this product to be positive, either both factors are positive, or both are negative.

If both are positive, then $x-9 > 0$ and $x-2 > 0$; from this you can see that $x >9$ and $x >2$. But the latter condition is redundant, since if $x >9$ then certainly $x >2$. Therefore, if both factors are positive, then $x >9$.

If both factors are negative, then following a similar chain of logic, you will get that $x <2$.

Thus either $x>9$ or $x<2$.

  • 1
    $\begingroup$ As an alternative to this algebraic approach, you could try a graphical approach. The left-hand side is a parabola with $x$-intercepts at $x=2$ and $x=9$. When is the parabola above the $y$-axis? $\endgroup$ – Théophile May 31 '14 at 13:43
  • $\begingroup$ So essentially, as you mentioned, if assumed x > 9 and x > 2, the latter is redundant and thus the larger number will always be the ">" sign? $\endgroup$ – mahat May 31 '14 at 13:47
  • $\begingroup$ @mahat Yes, that is correct. To make sure you understand this kind of reasoning, consider what happens if you reverse the inequality in the question: $(x-9)(x-2) < 0$. Then one product must be positive and one must be negative; try working out the logic from there. $\endgroup$ – Théophile May 31 '14 at 13:53

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