Let α and β be the roots of the quadratic equation 2x² - 11x + 5 = 0
∴ α + β = -(-11)/2 = 11/2 ...(i)
and αβ = 5/2
Now, (α - β)² = (α + β)² - 4αβ

Correct Option: A

Let α and β be the roots of the quadratic equation 2x² - 11x + 5 = 0
∴ α + β = -(-11)/2 = 11/2 ...(i)
and αβ = 5/2
Now, (α - β)² = (α + β)² - 4αβ
= (11/2)² - 4(5/2) = 121/4 - 20/2
= 121 - 40/4 = 81/4 = (9/2)²
∴ Difference of roots = (α - β) = 9/2 = 4.5

Let α = 2, β = 1/4
Then, α + β = 2 + 1/4 = 9/4
α β = 2.1/4 = 1/2
Equation having the roots α and β is
x² - (α + β)x + α β = 0

Correct Option: A

Let α = 2, β = 1/4
Then, α + β = 2 + 1/4 = 9/4
αβ = 2.1/4 = 1/2
Equation having the roots α and β is
x² - (α + β)x + α β = 0
⇒ x² - 9/4x + 1/2 = 0
⇒ 4x² - 9x + 2 = 0

Since, p and q are the roots of the equation x² + px + q = 0
Then, p + q = - p
and pq = q
Now, pq = q

Correct Option: A

Since, p and q are the roots of the equation x² + px + q = 0
Then, p + q = - p
and pq = q
Now, pq = q
⇒ p = 1
Putting the value of p in p + q = - p, we get
1 + q = - 1
⇒ q = - 2

We know that, AM ≥ GM
∴ √a + 1/√a ≥ 2
Here, √x² - x + 1 + 1/√x² - x + 1 ≥ 2
⇒ 2 - x² ≥ 2

Correct Option: B

We know that, AM ≥ GM
∴ √a + 1/√a ≥ 2
Here, √x² - x + 1 + 1/√x² - x + 1 ≥ 2
⇒ 2 - x² ≥ 2
⇒ x² ≤ 0
⇒ x = 0
Hence, the given equation has only one solution.

Let the roots be α and β .
Then, α + β = 8 ......(i)
α - β = 4 ..(ii)
On solving Eqs. (i) and (ii), we get
α = 6, β = 2
∴ Required equation is
x² - (α + β) x + (α β) = 0

Correct Option: D

Let the roots be α and β .
Then, α + β = 8 ......(i)
α - β = 4 ..(ii)
On solving Eqs. (i) and (ii), we get
α = 6, β = 2
∴ Required equation is
x² - (α + β) x + (α β) = 0
⇒ x² - (6 + 2)x + 6 x 2 = 0
⇒ x² - 8x + 12 = 0