Question

If one root of x2 + kx + 27 = 0 may be triple
the other, then K=
1) +6 2) +8 3) +12 4) +10​

Answer 1

Answer:

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Answer 2

Answer:

According to the question given the equation,

$x^2 + kx + 27 =0$

Given, one root is three times the other one

Let,

one root is $a$ so other one is $3a$

now we know that,

$a + 3a = - k \implies 4a = -k \: .....(*)$

now again,

we have,

$a \times 3a = 27 \implies 3a^2 = 27 \implies a^2 = 9\\\implies a = \pm 3$

Using equation $(*)$

for, $a = 3 , k = -12$

and for $a = -3 , k =12$

it matches with the option $(3)$

Hence, k = 12 (option 3)