Question

Find k so that one root of the equation 2x^2-16x+k=0 is twice the other

Answer 1

let one root be =ß
then,, as we know that sum of roots =-b/a
ß+2ß=-(-16)/2
3ß=8
ß=8/3
and product of roots =c/a
2ß^2=k/2
(8/3)^2=k/4
64/9*4=k
k=256/9

Answer 2

Answer:

The value of k is 258/9

Step-by-step explanation:

Quadratic Equation:

• The degree of the Quadratic Equation is two
• general form - $ax^{2} + bx + c = 0$
• The values which satisfy the quadratic equation are called roots

for a given Quadratic equation

the sum of roots is given by  = - coefficient of x / coefficient of x² = (-b/a)

the product of roots is given by = constant / coefficient of x² = (c/a)

let us consider one root as α

given one root is twice the other so,

sum of roots =

α + 2α = - ((-16) / 2)

3α = - ( -8)

3α = 8

α = 8/3

∴ roots are 8/3 and 16/3

product of roots =

α × 2α = (k/2)

2α² = k/2

k = 4α²

k = 4 × (8/3)²

k = (4 × 64) / 9

k = 256 / 9

hence, the value of k is 256/9

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