# The inequality x^4 < 8x^2 + 9 is satisfied precisely when...

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This is a multiple choice question, with possible answers:

(a) -3 < x <  3;

(b)  0 < x <  4;

(c)  1 < x <  3;

(d) -1 < x <  9;

(e) -3 < x < -1.

Let's start by rearranging the inequality to get

x^4 - 8x^2 - 9 < 0.

Now, we notice that x^4 = (x^2)^2, and so what we have on the left-hand side of our inequality above is really a quadratic equation in x^2:

(x^2)^2 - 8x^2 - 9 < 0.

So we can factor this like a normal quadratic: look for two numbers that add to make -8 and multiply to make -9. It turns out that -9 and +1 work, then our inequality is simply

(x^2 - 9)(x^2 + 1) < 0.

If we multiply two numbers together, the only way for the product to be negative (less than zero) is for one of the numbers to be negative and the other positive. But x^2 is positive no matter what value x takes, and so (x^2 + 1) is definitely going to be positive, for all values of x.

So we have to have (x^2 - 9) negative:

x^2 - 9 < 0.

And so

x^2 < 9,

hence

-3 < x < 3,