Question

11. If one root of the equation 2x2 -ax +64 = 0 is twice the other, then find the value of a

Answer 1

Let one root of given equation 2x² - ax + 64 = 0 is r
Then, other root is 2r .
Now, sum of roots = -coefficient of x/coefficient of x²
r + 2r = -(-a)/2
3r = a/2
a = 6r -----(1)

product of roots = Constant/coefficient of x²
r.2r = 64/2
2r² = 32
r² = 16
Taking square root both sides,
r = ±4
hence, roots are 4 , 8 or -4 , -8

if r = 4 then, a = 6r = 24
if r = -4 then, a = 6r = -24

Answer 2

HELLO DEAR,

let one of the zeroes of the Equation 2x² - ax + 64 = 0 be x , other be 2x

x + 2x = -(b/a)

3x = -(-a/2)

a = 6x------------( 1 )


x*2x = (c/a)

2x² = 64/2

x² = 32/2 = 16

x = +-4

If x = 4 , then other root is 2(4) = 8,
if x = -4 , then other root is 2(-4) = -8



now,

if x = 4, then a = 6(4) = 24
if x = -4 , then a = 6(-4) = -24



I HOPE ITS HELP YOU DEAR,
THANKS