Question

Find the value of p, for which one zero of the quadratic polynomial px² – 14x + 8 is 6 times the other.

Answer 1

Given

`{:(px^2 - 14x + 8),(ax^2 + bx + c):}}` On comparison

a = p, b = –14, c = 8

Given that one zero of the given poly is 6 times the other.

Let α and β be two roots.

ATQ, β = 6α

Now, `α + β = (-b)/a`

`α + 6α = (-(-14))/p`

`7α = (14)/p`

`α = 14/(p xx 7)`

`α = 2/p`

`αβ = c/a`

`α(6α) = 8/p`

`6α^2 = 8/p`

`6 (2/p)^2 = 8/p`

`6 (4/p^2) = 8/p`

`(24)/p^2 = 8/p`

`3/p = 1` 

⇒ p = 3