Question

if one zero of the polinomisl2x^2-5x-(2k+1) is twice the other then find both the
zeroes of the polynomial and value of k.​

Answer 1

heya mate there's your answer

if ax^2+bx+c=0 is the quadratic equation

then ,•sum of roots =-b/a

•product of roots =c/a

So on comparing given polynomial

2x^2-5x-(2k+1)=0 with this we get,

a=2, b=-5 ,c=-(2k+1)

so sum of roots =-(-5/2)=5/2

product of roots =-(2k+1)/2

but given one root is twice the other

so let one root be alpha and another be 2alpha

Now, alpha+2alpha =5/2

3alpha=5/2

alpha =5/6

Now,

alpha ×2alpha =-(2k+1)/2

2alpha square =-(2k+1)/2

alpha square=-(2k+1)/4

we got alpha =5/6 so,

(5/6)^2=-(2k+1)/4

25/36=-(2k+1)/4

25/9=-(2k+1)

25/9=-2k-1

2k=-1-25/9

2k=-9-25/9

2k=-34/9

k=-17/9

hope it helps you

Thank you