1. Let α, β be the roots of the equation x2 - px + r = 0 and α/2, 2β be the roots of the equation x2 - qx + r = 0. Then the value of r is ?
"Let α, β be the roots of the equation x2 - px + r = 0"
This implies:
(x-α)(x-β)=0
or
r=αβ ...(1)
p=α+β ...(2)
"and α/2, 2β be the roots of the equation x2 - qx + r = 0."
implies:
(x-α/2)(x-2β)=0
or
r=αβ ...(1a)
q=α/2+2β ...(3)
From (2) and (3) solve for α and β in terms of p and q, hence find r=αβ
To find the value of r, we can use the fact that the sum of the roots of a quadratic equation is equal to the negation of the coefficient of the linear term (p) divided by the coefficient of the quadratic term (1 in this case).
1. Let's start with the equation x^2 - px + r = 0. The sum of the roots, α and β, is (-p)/1 = -p.
2. Now consider the equation x^2 - qx + r = 0. The sum of the roots, α/2 and 2β, is (-q)/1 = -q.
3. We know that α/2 and 2β are the roots of x^2 - qx + r = 0. Therefore, the sum of the roots, α/2 + 2β, is equal to -q.
4. From equations (2) and (3), we have -p = -q.
5. Since -p = -q, we can conclude that p = q.
6. Using equation (1), which states that the sum of the roots of x^2 - px + r = 0 is -p, we can say that α + β = -p.
7. Similarly, since α/2 + 2β is the sum of the roots of x^2 - qx + r = 0, we have α/2 + 2β = -q = -p.
8. Rearranging equation (7), we get α + 4β = -2p.
9. Subtracting equation (6) from equation (8), we have 3β = -p.
10. From equation (10), we can conclude that β = -p/3.
11. Now, let's substitute the value of β from equation (11) into equation (6): α + (-p/3) = -p.
12. Solving equation (12) for α, we get α = (-2p)/3.
13. Finally, we can substitute the values of α and β into one of the equations (1 or 2) to find the value of r. Let's use equation (1): (-2p/3)(-p/3) + r = 0.
14. Simplifying equation (13), we get (2p^2)/9 + r = 0.
15. Rearranging equation (14) to solve for r, we have r = -(2p^2)/9.
Therefore, the value of r is -(2p^2)/9.