Question

Quadatric equation is given by (2a+1)x^2-ax+a-1=0, and two real roots are p and q.
Find the value range of a such that p >1 and q <1.

Answer 1

Do you have the answers? If so, could you show me to see I got the same answer.

Answer 2

@Azteck no i don't know how to solve this :/

Answer 3

\[ax^2-bx+c=0\]
\[\alpha + \beta=\frac{ -b }{ a }\]
\[\alpha \beta=\frac{ c }{ a }\]
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\[p+q=\frac{-(-a)}{2a+1}\]
\[pq=\frac{ a-1 }{ 2a+1 }\]
DISCRIMINANT
\[b^2-4ac \ge 0\]
^real roots means greater than or equal to zero.
\[(-a)^2-4(2a+1)(a-1) \ge 0\]
\[a^2-4(2a^2-a-1) \ge 0\]
\[a^2-8a^2+4a+4 \ge 0\]
\[-7a^2+4a+4 \ge 0\]
\[7a^2-4a-4 \le 0\]

Answer 4

I think you need to use the discriminant plus the sum of roots.

Answer 5

and product of roots.

Answer 6

@Azteck Thank u so much for ur answer. It's really helpful:)

Answer 7

But at one point I'm still in doubt. How can I make sure the two roots are satisfying this condition: p >1 and q <1

Answer 8

I'm not entirely sure. I'm trying to think.

Answer 9

Do you have the correct answers. Do you have an answer sheet or something like that?

Answer 10

Nope, unfortunately I have only the questions sheets :/

Answer 11

I don't know if the p and q as roots are just extraneous information. I think I need assistance. @satellite73 Do you know how to answer this question?

Answer 12

@jim_thompson5910

Answer 13

firstly, let us go for real roots,
as @azteck rightly did,
we have
7a^2 - 4a - 4 <= 0
Note we don't want equal roots, hence
7a^2 - 4a - 4 < 0

Let us try to say from the graph,
If, 2a+1 >0, the graph will be an upward parabola like this :

If 2a-1 <0, the graph will have be a downward parabola, like this:

Where we have used the info that 1 lies between both roots,
from either of the graphs, you can easily see that our 2nd required condition would be
(2a+1)(f(1)) <0

=> (2a+1)(2a + 1 - a +a -1 )<0
=>(2a+1)(a) <0

Solve this inequality, and the previous one, and take the common points as the final answer.
Hope that helps.