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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I would start this way: $\displaystyle x^2 - 4x - 5 < 0$ $\displaystyle (x - 5)(x + 1) < 0$ Can you finish from here?
Originally Posted by icemanfan I would start this way: $\displaystyle x^2 - 4x - 5 < 0$ $\displaystyle (x - 5)(x + 1) < 0$ Can you finish from here? ah yees yes i understand what u did. yeah i can finish it. thank u!
Originally Posted by icemanfan I would start this way: $\displaystyle x^2 - 4x - 5 < 0$ $\displaystyle (x - 5)(x + 1) < 0$ Can you finish from here? x < 5 , x < -1 is that it ?
Last edited by jvignacio; Oct 2nd 2008 at 08:42 PM.
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.
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 ????
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.
Not necessary to make a sketch, just complete the square: $\displaystyle x^{2}-4x-5<0\implies (x-2)^{2}<9\implies -3<x-2<3,$ and we're done.
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