Question

Given two quadratic equations,

I. x2 - 31x + 240 = 0

Equation I has two roots x1, x2. (x1 > x2)

II. 3y² – 7y – 3 = 8y - 3y²- 9

Equation II has two roots y₁, y₂. (y₂ > y₁)

Find the equation whose roots are (– 2y₁ + x₁), - (y₂ – x₂).

1. x² - 28x - 195 = 0
2. x2 - 28x + 195 = 0
3. x2 + 28x - 195 = 0
4. x2 + 28x + 195 = 0
5. Cannot be determined

Answer 1

Correct Answer - Option 2 : x2 - 28x + 195 = 0

I. x² - 31x + 240 = 0

⇒ x² - 16x - 15x + 240 = 0

⇒ x(x - 16) - 15(x - 16) = 0

⇒ (x - 16) (x - 15) = 0

⇒ x = 15 , 16

So, According to the condition x₁ = 16 & x₂ = 15

II. 3y² – 7y – 3 = 8y - 3y²- 9

⇒ 6y² – 15y + 6 = 0

⇒ 6y² – 12y – 3y + 6 = 0

⇒ 6y(y – 2) – 3(y - 2) = 0

⇒ (y – 2)(6y – 3) = 0

⇒ y = 2 or y = 1/2

So, According to the condition y₁ = 1/2 & y₂ = 2

(– 2y₁ + x₁) = (– 1 + 16) = 15

- (y₂ – x₂) = - (2 - 15) = 13

So, the equation whose roots are 13 & 15 is k(x - 15)(x - 13), (where k can be any number)

Equation is: x² - 28x + 195 = 0