$\displaystyle -4\sqrt{x}+4>12$

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- Dec 9th 2009, 03:29 PM #1

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- Dec 9th 2009, 03:40 PM #2

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- Dec 9th 2009, 03:45 PM #3

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No, that isn't correct.

#1, when dividing by -4 the inequality has to flip.

#2 the answer has to be a range, not just one number.

The answer is x>4, but I don't know how to arrive at the answer after:

$\displaystyle \sqrt{x}<-2$

x<4 (how do we get to x>4?)

- Dec 9th 2009, 03:46 PM #4

- Dec 9th 2009, 03:47 PM #5
Wrong. If you multiply or divide by a negative number, it changes the direction of the inequality.

First, it is important to note that $\displaystyle x \geq 0$, since you can not have the square root of a negative number.

$\displaystyle -4\sqrt{x} + 4 > 12$

$\displaystyle -4\sqrt{x} > 8$

$\displaystyle \sqrt{x} < -2$.

$\displaystyle x < (-2)^2$

$\displaystyle x < 4$.

Putting it all together, we have our solution set

$\displaystyle 0 \leq x < 4$.

Notice that if you use any number less than 4, you need to use the NEGATIVE square root to make the sentence make sense...

- Dec 9th 2009, 03:48 PM #6

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That isn't a correct solution set.

Substitute some values back into the inequality. For example if x=1, the inequality isn't true.

-4(-1)+4=8 which isn't greater than 12.

-4(1)+4=0 which isn't greater than 12.

The answer is x>4, but how do we arrive at that?

- Dec 9th 2009, 03:48 PM #7

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- Dec 9th 2009, 04:01 PM #8

- Dec 9th 2009, 04:02 PM #9

- Dec 9th 2009, 04:16 PM #10

- Dec 9th 2009, 05:13 PM #11
There is no solution since the inequality reduces to $\displaystyle \sqrt{x} < -2$ and the left hand side is always greater than or equal to zero.

Not okay. For example, x = 25 lies in this 'solution'. Does it satisfy the original equation ....?

- Dec 9th 2009, 05:19 PM #12

- Dec 9th 2009, 05:28 PM #13

- Dec 9th 2009, 05:42 PM #14

- Dec 9th 2009, 06:14 PM #15