How Is This Solved

**fcabanski** · December 9th 2009, 03:29 PM

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

**jgv115** · December 9th 2009, 03:40 PM

I hope I get this right:

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

$-4\sqrt{x}>8$

$\sqrt{x}<-2$

$x < (-2)^2$

$x< 4$

**fcabanski** · December 9th 2009, 03:45 PM

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:

$\sqrt{x}<-2$
x<4 (how do we get to x>4?)

**pickslides** · December 9th 2009, 03:46 PM

Originally Posted by jgv115

$-4\sqrt{x}>8$

$\sqrt{x}>-2$

dividing by a negative number will change the inequality.

therefore

$-4\sqrt{x}>8$

$\sqrt{x}<-2$

**Prove It** · December 9th 2009, 03:47 PM

Originally Posted by jgv115

I hope I get this right:

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

$-4\sqrt{x}>8$

$\sqrt{x}>-2$

$x > (-2)^2$

$x= 4$

Wrong. If you multiply or divide by a negative number, it changes the direction of the inequality.

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

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

$-4\sqrt{x} > 8$

$\sqrt{x} < -2$ .

$x < (-2)^2$

$x < 4$ .

Putting it all together, we have our solution set

$0 \leq x < 4$ .

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

**fcabanski** · December 9th 2009, 03:48 PM

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?

**jgv115** · December 9th 2009, 03:48 PM

Yea, i know..

sorry about the mistake. I did it right on paper but just didn't type it right.

**Bacterius** · December 9th 2009, 04:01 PM

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

$-4\sqrt{x}>8$

$\sqrt{x}<-2$

$-1 \sqrt{x}> 2$

$(-1 \sqrt{x})^2> (2)^2$

$(-1)^2 \sqrt{x}^2> 2^2$

$1 x > 4$

$x > 4$

Okay ?

**Prove It** · December 9th 2009, 04:02 PM

Much better

**fcabanski** · December 9th 2009, 04:16 PM

Thanks.

**mr fantastic** · December 9th 2009, 05:13 PM

Originally Posted by fcabanski

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

There is no solution since the inequality reduces to $\sqrt{x} < -2$ and the left hand side is always greater than or equal to zero.

Originally Posted by Bacterius

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

$-4\sqrt{x}>8$

$\sqrt{x}<-2$

$-1 \sqrt{x}> 2$

$(-1 \sqrt{x})^2> (2)^2$

$(-1)^2 \sqrt{x}^2> 2^2$

$1 x > 4$

$x > 4$

Okay ?

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

**VonNemo19** · December 9th 2009, 05:19 PM

Originally Posted by mr fantastic

There is no solution since the inequality reduces to $\sqrt{x} < -2$ and the left hand side is always greater than or equal to zero.

I was wondering when someone was gonna point that out.

**Prove It** · December 9th 2009, 05:28 PM

Originally Posted by VonNemo19

I was wondering when someone was gonna point that out.

I was thinking that, but then I thought "every square root has two solutions, one positive and one negative".

E.g. $\sqrt{9} = 3$ or $\sqrt{9} = -3$ .

And $-3 < -2$ ...

**mr fantastic** · December 9th 2009, 05:42 PM

Originally Posted by Prove It

I was thinking that, but then I thought "every square root has two solutions, one positive and one negative".

E.g. $\sqrt{9} = 3$ or $\sqrt{9} = -3$ .

And $-3 < -2$ ...

No. $\sqrt{9} = 3$ . $-\sqrt{9} = -3$ .

If $x^2 = 9$ then it's true to say that $x = \pm \sqrt{9} = \pm 3$ . But the notation $\sqrt{9}$ means the positive square root of 9.

**Stroodle** · December 9th 2009, 06:14 PM

Originally Posted by fcabanski

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

It looks to me like there is no solution as:

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

$4\sqrt{x}+8<0$

There are no real values of $x$ that satisfy this inequality.

Thread: How Is This Solved

LinkBack

Thread Tools

Display

How Is This Solved

Search Tags