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21 Feb 2009, 20:17
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What are the values of X that satisfy the inequality x+1/x < 2.5?
Ans :
1/2 < x < 2 .
But i felt x can also be any negative value.
Explain please.
Ans :
1/2 < x < 2 .
But i felt x can also be any negative value.
Explain please.
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21 Feb 2009, 20:49
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tkarthi4u,
Please go thru this link. Hope you will get some information on how to solve the Quadratic Inequalities.
https://www.purplemath.com/modules/ineqquad.htm
Please go thru this link. Hope you will get some information on how to solve the Quadratic Inequalities.
https://www.purplemath.com/modules/ineqquad.htm
10 Mar 2009, 10:49
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Hey mrsmarthi, i went through the link given for quad. eqns. It was of tremendous help.Thanks !!
I am posting my solution to the problem in this post using that method. Can you ( or anyone else perhaps) review it and tell me if it is indeed, the right way to go.
SOLUTION :
Since x+1/x<2.5,
I consider x+1/x-2.5<0 and further, i break it down to x+1/x-2.5 = y and y<0.
Now I solve the quadratic eqn considering y=0 i.e. x+1/x-2.5=0 and get the solution, (x-2)*(x-.5)=0.
Thus, on the number line, I plot the points .5 and 2 on the x -axis. We now have three regions on the x axis, which are
x>2, .5<x<2 and x<.5.
Now, for the main equation, we need to see y<0 lies in which of the three regions.
x+1/x-2.5 =0 can be written as x^2-2.5x+1=0. We can find the vertex of the parabola using the formula, v=(-b/2a,c-b^2/4a).
Which gives us v below the x -axis between .5 and 2 and .5 and 2 are the x-intercepts of the parabola.
Thus, between .5 and 2, the parabola x+1/x-2.5 or simply y is less than zero.
I am posting my solution to the problem in this post using that method. Can you ( or anyone else perhaps) review it and tell me if it is indeed, the right way to go.
SOLUTION :
Since x+1/x<2.5,
I consider x+1/x-2.5<0 and further, i break it down to x+1/x-2.5 = y and y<0.
Now I solve the quadratic eqn considering y=0 i.e. x+1/x-2.5=0 and get the solution, (x-2)*(x-.5)=0.
Thus, on the number line, I plot the points .5 and 2 on the x -axis. We now have three regions on the x axis, which are
x>2, .5<x<2 and x<.5.
Now, for the main equation, we need to see y<0 lies in which of the three regions.
x+1/x-2.5 =0 can be written as x^2-2.5x+1=0. We can find the vertex of the parabola using the formula, v=(-b/2a,c-b^2/4a).
Which gives us v below the x -axis between .5 and 2 and .5 and 2 are the x-intercepts of the parabola.
Thus, between .5 and 2, the parabola x+1/x-2.5 or simply y is less than zero.
