Question

if one of roots of 2x^2+ax+32=0 is twice the other root,then the value of a is​

Answer 1

Step-by-step explanation:

so let one root be M

and another will be 2M

so ,

f (M) = 2 (M)^2+a(M)+32 =0

a = (-32 -2M^2)/M

f (2M) = 2(2M)^2 +a(2M) +32 = 0

a = (-8M^2 -32)/2M

now we have 2 values of a

2(-32-2M^2) = -8M^2 -32

-64 -4M^2 = -8M^2 -32

4M^2 = 64 -32 = 32

M^2 = 32/4 = 8

M = 2root2

now value of a will be

= -32-2 (2root2)^2/ (2root2)

= -48/2root2

= -24/ root2 = 12 (root2)^2/root2

= 12root2 answer

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