Inequalities

[stable gulch]

Hi how would I do question 8?

[zenith tundra]

Start by cross multiplying
$$\frac{2}{x^2} \ge \frac{3}{(x + 1)(x - 2)}$$
$$\implies 2(x + 1)(x - 2) \ge 3\cdot x^2$$
Then from here I believe you know how to solve this considering you have solved questions 1 through 7

[white zealot]

HitenTandon

[mellow kiln]

You have to be careful with cross multiplying, as multiplying a side by a negative number changes the direction of the inequality sign.

To handle that, you need to find out when the thing you're multiplying both sides by is negative and when it's not and handle those separately.

So, x^2 will always be nonnegative, so that's not an issue.

But (x + 1)(x - 2) might be negative sometimes.

So, you get the roots, -1 and 2.

[limber plaza]

You are right!

[mellow kiln]

Then, you see the sign of (x + 1)(x - 2) when it's left of -1, when it's between -1 and 2, and when it's right of 2.

We can use -2 as an example of a number that's left of -1.

(x + 1)(x - 2) = ((-2) + 1)((-2) - 2) = (-1)(-4) = 4, so left of -1, it's positive.

We can use 0 as an example of a number between -1 and 2.

(x + 1)(x - 2) = (0 + 1)(0 - 2) = (1)(-2) = -2, so in between -1 and 2, it's negative.

We can use 3 as an example of a number right of 2.

(x + 1)(x - 2) = (3 + 1)(3 - 2) = (4)(1) = 4, so right of 2, it's positive.

[limber plaza]

Great!!! The answer is great but the instruction is more helpful.

[mellow kiln]

So, now we can do the cross multiplication with two cases.

[white zealot]

Chai T. Rex

[mellow kiln]

So, you can now solve the top one for x.

Throw out any solutions that are forbidden by the 'when' condition.

Then you can solve the bottom one for x.

Again, throw out any solutions that are forbidden by the 'when' condition.

[stable gulch]

Oh I see thanks so much! My book was using some different method for other questions 😔