Question

Find the value of k for which the roots a and b of equation : x^-6x+k=0 satisfy the relation 2a+3b=20.

Answer 1

Answer:

Step-by-step explanation:

P(x)=x²-6x+k

And 'a' and 'b' are the roots

So, a+b=-(coefficient of x)/(coefficient of x²)

or a+b=-(-6)/1

or a+b=6 eq. (1)

It is given that

2a+3b=20 eq.(2)

Multiplying eq.(1) by 2 we get,

2a+2b=12 eq.(3)

Subtracting (3) from (2) we get

2a+3b-2a-2b=20-12

b=8

Putting b=8 in (1)

a+8=6

a=6-8

a=-2

Now,

ab = constant term/coefficient of x²

8×-2=k/1

-16=k

K=-16

hope this help you

pls mark my answer as Branliest

Answer 2

Answer:

P(x)=x²-6x+k

And 'a' and 'b' are the roots

So, a+b=-(coefficient of x)/(coefficient of x²)

or a+b=-(-6)/1

or a+b=6 ......eq. (1)

It is given that

2a+3b=20 eq.(2)

Multiplying eq.(1) by 2 we get,

2a+2b=12 eq.(3)

Subtracting (3) from (2) we get

2a+3b-2a-2b=20-12

b=8

Putting b=8 in (1)

a+8=6

a=6-8

a=-2

Now,

ab = constant term/coefficient of x²

8×-2=k/1

-16=k

K=-16