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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
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B
-29
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C
19
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D
15
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Solution

The correct option is A 29
let 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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