Question

find the value of p for which one root of the quadratic equation px2 -14x +8=0 is 6 times the other

Answer 1

Given -

px² - 14x + 8 = 0

Let one of the roots be 'a' , then the other root is '6a'

a + 6a = 14/p
7a = 14/p
a = 2/p ---(i)

a x 6a = 8/p
6a² = 8/p

Plugging the value of a from eq (i),
6 x (2/p)² = 8/p
6 x 4/p² = 8/p
24/p = 8   (Cancelling 'p' from both the sides)
8p = 24
p = 3

So required value of 'p' is 3

Hope This Helps You

Answer 2

px^2-14x+8=0

let  a and b be the roots of the equation

b=6a

sum of roots = -b/a

product of roots = c/a

here a=p , b=-14 , c=8

a+b=14/p

ab=8/p

a+6a=14/p

7a=14/p

a=2/p

a*6a=8/p

6a^2=8/p

3a^2=4/p

3*(2/p)^2=4/p

3*4/p^2=4/p

p=3