1. How Is This Solved

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

2. I hope I get this right:

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

$-4\sqrt{x}>8$

$\sqrt{x}<-2$

$x < (-2)^2$

$x< 4$

3. 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?)

4. Originally Posted by jgv115

$-4\sqrt{x}>8$

$\sqrt{x}>-2$
dividing by a negative number will change the inequality.

therefore

$-4\sqrt{x}>8$

$\sqrt{x}<-2$

5. 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...

6. 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?

7. Yea, i know..

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

8. $-4\sqrt{x}+4>12$

$-4\sqrt{x}>8$

$\sqrt{x}<-2$

$-1 \sqrt{x}> 2$

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

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

$1 x > 4$

$x > 4$

Okay ?

9. Much better

10. Thanks.

11. 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 ....?

12. 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.

13. 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 $-3 < -2$...

14. 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.

15. 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.

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