Question

please answer ..........***---******

Answer 1

let α , β be the roots of quadratic equation x² - 6x + a = 0

⇒ then sum of the two roots (α, β) : α + β = 6

⇒ product of the two roots (α, β) : α × β = a

given that one of two roots of quadratic equation x² - 6x + a = 0 is common to the quadratic equation x² - bx + 6 = 0

let α be the common root to the both quadratic equations and let Δ be the other root of the quadratic equation x² - bx + 6 = 0

⇒ sum of the two roots (α, Δ) : α + Δ = b

⇒ product of the two roots (α , Δ) : α × Δ = 6

let us divide the product of the roots of 1st quadratic equation with the product of roots of the 2nd quadratic equation

⇒ (α × β)/(α × Δ) = a/6

⇒ β/Δ = a/6

but given that the other root of 1st quadratic equation and 2nd quadratic equation are in the ration of 4 : 3

⇒ β/Δ = 4/3

but we know that β/Δ = a/6

⇒ a/6 = 4/3

⇒ a = 8

⇒ the 1st quadratic equation is x² - 6x + 8 = 0

⇒ x² - 4x -2x + 8 = 0

⇒ x = 4 or x = 2

we chose that α and β are the roots of 1st quadratic equation, but we cannot decide which value is α and which value is β, (i.e.) α can be 4 and α can be 2

# let us take the situation α = 4

we know that product of the two roots of 2nd quadratic equation (α , Δ) : α × Δ = 6

substituting the value of α = 4 in the above equation we get:

⇒ 4 × Δ  = 6

⇒ Δ = 1.5

it is mentioned clearly in the question that the other root of the 1st quadratic equation and 2nd quadratic equation are integers.

but we got Δ which is the other root of 2nd quadratic equation as 1.5 which clearly not an integer.

so we conclude from the above explanation that α cannot be 4.

⇒ we can conclude that α = 2

⇒ the common root of the two quadratic equations is 2

You got the required answer for the question, but am explaining the remaining for the sake of complete solution to the problem.

we know that product of the two roots of 2nd quadratic equation (α , Δ) : α × Δ = 6

we concluded that α = 2

⇒ 2 × Δ = 6

⇒ Δ = 3

we know that sum of the two roots of 2nd quadratic equation (α, Δ) : α + Δ = b

⇒ 2 + 3 = b

⇒ b = 5

⇒ the 2nd quadratic equation is x² - 5x + 6 = 0

⇒ the roots of quadratic equation x² - 6x + 8 = 0 are 4 and 2

⇒ the roots of quadratic equation x² - 5x + 6 = 0 are 3 and 2