Question

# If one  root of $$x^3-12x^2+kx-18=0$$ is thrice the sum fo the remaining two roots then $$k=$$

A
29
B
-29
C
19
D
15

Solution

## The correct option is A 29let roots are $$a,b,3(a+b)$$sum of roots $$-(-12)$$$$4(a+b)=12$$$$a+b=3$$product$$=-(-18)$$$$3ab(a+b)=+18$$$$ab=+2$$sum of two at$$=k$$$$ab+3a(a+b)+3b(a+b)=k$$$$k=2+9a+9b$$$$k=9\times 3+2$$$$k=29$$Mathematics

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