If one of the roots of quadratic equation X2 – kX + 27 = 0 is 3 then find
the value of ‘k’.
Answers
Answered by
1
Answer:
ax^2+bx+c=0
hence,a=1
b=k
c=27
hence,d=b^2-4ac
hence,d=k^2-4×1×27=k^2-108
d=3
hence,3=k^2-108
k^2=-105
pls mark me as brainliest
Answered by
2
Answer:
x^2-kx+27=0
a= 1 b= -k c= 27
α= 3 given
α+β= -b/a
3+β = k -------------------(i)
α*β = c/a
3*β= 27
β=9
substituting in (i)
k= 3+9= 12
Step-by-step explanation:
Similar questions