Question

find the value of p for which one root of the quadratic equation px^2-14x+8=0 is 6 times the other​

Answer 1

Answer:

p = 3

Step-by-step explanation:

let us take one root as x

and the other as 6x .

sum of zeros = x + 6x = 7x = -b/a = 14/p

=> p = 14/7x = 2/x ____________(i)

product of zeros = x × 6x = 6x² = c/a = 8/p

=> p = 8/6x² = 4/3x² ____________(ii)

from (i) and (ii) ,

2/x = 4/3 x²

on cross multiplication,

2×3x²=4×x = 6x² = 4x

this forms a quadratic equation,

6x²-4x+0 = 0

by middle term splitting,

sum = -4 = -4 + 0

product = 0 = -4 × 0

6x²- 4x + 0x + 0 = 0

2x ( 3x -2 ) + 0 ( 3x -2 ) = 0 [since (3x -2)×0 = 0]

(2x+ 0 ) (3x-2)

=> x= 0 or 2/3

since the root of the given polynomial will never be 0 , the root of the equation

x = 2/3

Substitute this in (i) ,

=> p = 2/x = 2/(2/3) = 3

Hope this will help you.