Roots lying between the roots of a given equation

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In summary, the required conditions for the roots of the equation x^2-2x-(a^2-1)=0 to lie between the roots of the equation x^2-2(a+1)x+a(a-1)=0 are that the discriminants of both equations must be non-negative, and the x-coordinate at which the minimum value of the function in the second equation occurs must be between the roots of the second equation. The range of values for 'a' that satisfy these conditions is a\in \left(\frac{-1}{4},1\right).
  • #1
utkarshakash
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Homework Statement


For what real values of 'a' do the roots of the equation [itex]x^2-2x-(a^2-1)=0[/itex] lie between the roots of the equation [itex]x^2-2(a+1)x+a(a-1)=0[/itex]

Homework Equations



The Attempt at a Solution


The required conditions are
[itex] \large D_1\geq0[/itex]
[itex] \large D_2\geq0[/itex]
[itex] \large R_1<\dfrac{-2}{2}<R_2[/itex]
where
[itex]D_n=[/itex]Discriminant of nth equation and [itex]R_m=[/itex] mth roots of the latter equation

Solving simultaneously the above inequalities and taking intersection
[itex]a\in \left(\frac{-1}{3},0\right)[/itex]
But the correct answer is
[itex]a\in \left(\frac{-1}{4},1\right)[/itex]
 
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  • #2
Where does the -2/2 come from? Shouldn't there be an 'a' in there?
 
  • #3
haruspex said:
Where does the -2/2 come from? Shouldn't there be an 'a' in there?

Sorry I made a mistake. It must be 2/2 (-b/2a)
 
  • #4
utkarshakash said:
Sorry I made a mistake. It must be 2/2 (-b/2a)
But that's only the average of the roots. You need the condition that each root individually is in that range.
 
  • #5
haruspex said:
But that's only the average of the roots. You need the condition that each root individually is in that range.

You are not understanding. Its not the average of the roots. Its the x-coordinate at which the minimum value of function occurs.
 
  • #6
utkarshakash said:
You are not understanding. Its not the average of the roots. Its the x-coordinate at which the minimum value of function occurs.
OK, but why is that interesting?
 
  • #7
Express the roots in terms of 'a' first. The non-negativity of the discriminants is not enough.

If you are right, both roots of the first equation lie between or equal to one of the roots of the second equation in case a=-1/3. Is it true?

ehild
 
  • #8
ehild said:
Express the roots in terms of 'a' first. The non-negativity of the discriminants is not enough.

If you are right, both roots of the first equation lie between or equal to one of the roots of the second equation in case a=-1/3. Is it true?

ehild

I started by finding roots then substituting them for x in the latter equation and setting them <0. The answer which I get is correct. So, no worries!
 
  • #9
Did you get [itex]a\in \left(\frac{-1}{4},1\right)[/itex]? That is the correct answer, I have checked. Why should the roots be negative? Try with a=0.5. Do the roots of the first equation lie between the roots of the second one?


ehild
 
  • #10
Yes i got that answer.
 

1. What does it mean for a root to lie between the roots of a given equation?

When we say that a root lies between the roots of a given equation, it means that the root is located somewhere in the interval between the two other roots of the equation. In other words, the root is a solution to the equation that falls between the other two solutions.

2. How can we determine if a root lies between the roots of a given equation?

To determine if a root lies between the roots of a given equation, we can use the Intermediate Value Theorem. This theorem states that if a continuous function takes on two values at two points in an interval, then it must take on every value between those two points. Therefore, if we can find two points in the interval that produce opposite signs when plugged into the equation, then we can conclude that a root must exist between those two points.

3. Can a root lie outside of the interval between the roots of a given equation?

Yes, it is possible for a root to lie outside of the interval between the roots of a given equation. This can occur if the function is not continuous or if there are other roots outside of the interval. However, if the function is continuous, the root must lie between the given interval.

4. How does finding roots between the roots of a given equation help in solving the equation?

Finding roots between the roots of a given equation can help in solving the equation by narrowing down the possible solutions. It allows us to focus on a smaller interval and use methods like the bisection method or Newton's method to approximate the root. This can make the process of solving the equation more efficient and accurate.

5. Are there any limitations to using the roots between the roots of a given equation?

Yes, there are limitations to using the roots between the roots of a given equation. This method can only be applied to continuous functions, and even then, it may not always provide an accurate solution. Additionally, if the function has multiple roots within the given interval, this method may not be able to determine which root is between the other two roots. It is important to consider these limitations when using this approach to solving equations.

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