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#1

$Old$ October 2nd, 2008, 12:03 PM

jvignacio $jvignacio is offline$

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$Default$ inequality help

x^2 - 4x < 5

= x - 4x < sqrt(5)

= -3x < sqrt(5)

= x > sqrt(5)/3

surely this is wrong!! any help thank u

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#2

$Old$ October 2nd, 2008, 12:13 PM

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I would start this way:

$x^2 - 4x - 5 < 0$

$(x - 5)(x + 1) < 0$

Can you finish from here?

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#3

$Old$ October 2nd, 2008, 12:16 PM

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Originally Posted by icemanfan $View Post$

I would start this way:

$x^2 - 4x - 5 < 0$

$(x - 5)(x + 1) < 0$

Can you finish from here?

ah yees yes i understand what u did. yeah i can finish it. thank u!

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#4

$Old$ October 2nd, 2008, 09:50 PM

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Originally Posted by icemanfan $View Post$

I would start this way:

$x^2 - 4x - 5 < 0$

$(x - 5)(x + 1) < 0$

Can you finish from here?

x < 5 , x < -1 is that it ?

Last edited by jvignacio; October 2nd, 2008 at 10:42 PM.

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#5

$Old$ October 2nd, 2008, 10:37 PM

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Originally Posted by jvignacio $View Post$

x < 5 , x <= -2 is that it ?

No. Draw the graph of y = (x - 5)(x + 1). For what values of x is y < 0 ....? That is, for what values of x is the graph below the x-axis ....?

Answer: -1 < x < 5.

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#6

$Old$ October 2nd, 2008, 11:01 PM

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No. Draw the graph of y = (x - 5)(x + 1). For what values of x is y < 0 ....? That is, for what values of x is the graph below the x-axis ....?

Answer: -1 < x < 5.

or x < 5 and x > -1 ????

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$Old$ October 2nd, 2008, 11:11 PM

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or x < 5 and x > -1 ????

I suppose so. But it's a poor way of stating the correct answer and could easily be misunderstood.

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#8

$Old$ October 3rd, 2008, 11:01 AM

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Not necessary to make a sketch, just complete the square:

$x^{2}-4x-5<0\implies (x-2)^{2}<9\implies -3<x-2<3,$ and we're done.

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