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Question

Find k if one root of the equation $$ \displaystyle x^{2}-6kx+8=0 $$  is twice the other 


A
2
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B
±1
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C
±12
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D
3
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Solution

The correct option is B $$ \displaystyle \pm 1 $$
since the one root is twice the others,
Let the one root be a and other root will be 2a
Now sum of roots $$a+2a=6k$$
and product of roots $$2a^2=8$$
$$\therefore a=2,-2$$

Now,
$$3a=6k$$
$$k= \frac{a}{2}$$
Now putting the obtained value of a, we get
$$k=1,-1$$

Maths

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