Given, one root of x² - 6kx + 5 = 0 is 5.
∴ x = 5 satisfies x² - 6kx + 5 = 0

Correct Option: C

Given, one root of x² - 6kx + 5 = 0 is 5.
∴ x = 5 satisfies x² - 6kx + 5 = 0
⇒ 5² - 30k + 5 = 0
⇒ 25 - 30k + 5 = 0
⇒ 30 - 30k = 0
⇒ 30k = 30
∴ k = 1

2x² -11x + 15 = 0
[by factorisation method]
⇒ 2x² - (6x + 5x) + 15 = 0

Correct Option: A

2x² -11x + 15 = 0
[by factorisation method]
⇒ 2x² - (6x + 5x) + 15 = 0
⇒ 2x² - 6x - 5x + 15 = 0
⇒ 2x(x - 3) - 5 (x - 3) = 0
⇒ (2x - 5) (x - 3) = 0
∴ x = 5/2, 3
Hence, the roots are 5/2 and 3.

[a + (1/a)]² = 3
Taking square roots both sides, we get
a + (1/a) = √3
On cubing both sides, we
[a + (1/a)]³ = (√3)³

Correct Option: B

[a + (1/a)]² = 3
Taking square roots both sides, we get
a + (1/a) = √3
On cubing both sides, we
[a + (1/a)]³ = (√3)³
⇒ a³ + 1/a³ + 3.a.1/[a(a + 1/a)] = 3 √3
⇒ a³ + 1/a³ + 3√3 = 3√3
∴ a³ + 1/a³ = 0

2x² - 7xy + 3y² = 0
⇒ 2x² - 6xy - xy + 3y² = 0
⇒ 2x(x - 3y) - y(x - 3y) = 0
⇒ (2x - y) (x - 3y) = 0

Correct Option: C

2x² - 7xy + 3y² = 0
⇒ 2x² - 6xy - xy + 3y² = 0
⇒ 2x(x - 3y) - y(x - 3y) = 0
⇒ (2x - y) (x - 3y) = 0
Either, 2x - y = 0 ⇒ 2x = y
⇒ x/y = 1/2
or x - 3y = 0
⇒ x = 3y ⇒ x/y = 3/1

Given quadric equation is
x² + x/b + 1/c = 0 ...(i)
Now, by condition the roots of the Eq.(i) are α and 1/α.
Now,product of roots = (1/c) / (1/a)
⇒ α.(1/α) = a/c

Correct Option: D

Given quadric equation is
x² + x/b + 1/c = 0 ...(i)
Now, by condition the roots of the Eq.(i) are α and 1/α.
Now,product of roots = (1/c) / (1/a)
⇒ α.(1/α) = a/c
⇒ c = a
which is the required relation.