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For how many values of a greater than 0

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For how many values of a greater than 0  [#permalink]

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New post 07 Jul 2020, 08:44
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For how many values of a greater than 0 (A>0), both the roots of ax^2 - (a+1)x + (a-2) = 0, are greater than 3?

A. 1
B. 0
C. 4
D. 5
E. 2

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Re: For how many values of a greater than 0  [#permalink]

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New post 07 Jul 2020, 10:36
1
Since both roots are greater than 3, sum of roots must be greater than 6.

\(\frac{a+1}{a} >6\)

\(a<\frac{1}{5 }\).......(1)

Also,

\(y= ax^2 - (a+1)x + (a-2\)) is an upward parabola, since a>0.
Attachment:
Untitled.png
Untitled.png [ 5.33 KiB | Viewed 370 times ]

f(3) > 0, since both roots are greater than 3.

9a-3(a+1)+a-2 >0

\(a> \frac{5}{7}\).......(2)

We can see that equation (1) and (2) both are contradictory. No such value of a is possible.

B





yashikaaggarwal wrote:
For how many values of a greater than 0 (A>0), both the roots of ax^2 - (a+1)x + (a-2) = 0, are greater than 3?

A. 1
B. 0
C. 4
D. 5
E. 2

Posted from my mobile device
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Re: For how many values of a greater than 0  [#permalink]

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New post 07 Jul 2020, 11:26
1
yashikaaggarwal wrote:
For how many values of a greater than 0 (A>0), both the roots of ax^2 - (a+1)x + (a-2) = 0, are greater than 3?

A. 1
B. 0
C. 4
D. 5
E. 2

Posted from my mobile device


Product of the roots = (a-2)/a
Sum of the roots = - (a+1)/a
We have to find when, -a -1 / a > 6 or, -a -1 > 6a or, -1 > 6a, since a is greater than 1. there is no such value.
and (a-2)/a > 9 or, 9a< a-2, or, 8a < -2, no value of a is possible.

B is the answer.
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Re: For how many values of a greater than 0  [#permalink]

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New post 07 Jul 2020, 12:20
minustark wrote:
yashikaaggarwal wrote:
For how many values of a greater than 0 (A>0), both the roots of ax^2 - (a+1)x + (a-2) = 0, are greater than 3?

A. 1
B. 0
C. 4
D. 5
E. 2

Posted from my mobile device

Product of the roots = (a-2)/a
Sum of the roots = - (a+1)/a
We have to find when, -a -1 / a > 6 or, -a -1 > 6a or, -1 > 6a, since a is greater than 1. there is no such value.
and (a-2)/a > 9 or, 9a< a-2, or, 8a < -2, no value of a is possible.

Sum of the roots would be (a+1)/a
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Re: For how many values of a greater than 0  [#permalink]

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New post 09 Jul 2020, 22:17
1
ax^2 - (a+1)x + (a-2) = 0 and both roots are greater than 3.

product of roots = c/a = (a-2)/a

Since each root is greater than 3, thus the product of the roots will be greater than 9.
=> (a-2)/a > 9
=> a-2 > 9a
=> a < -1/4

but a is greater than 0 (stated in question) => no such value of 'a' for which both roots are greater than 3.

Ans: B
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Re: For how many values of a greater than 0   [#permalink] 09 Jul 2020, 22:17

For how many values of a greater than 0

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