1. ## 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

2. I would start this way:

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

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

Can you finish from here?

3. Originally Posted by icemanfan
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!

4. Originally Posted by icemanfan
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 ?

5. Originally Posted by jvignacio
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.

6. Originally Posted by mr fantastic
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 ????

7. Originally Posted by jvignacio
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.

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

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