Question

one root of x^2-9x+k=0 is twice the other.

Answer 1

Step-by-step explanation:

Let one root be y

another root = 2y

$sum \: of \: roots = \frac{ - b}{a} \\ y + 2y = \frac{ - ( - 9)}{1} \\ 3y = 9 \\ y = \frac{9}{3} \\ y = 3$

$product \: of \: roots \: = \frac{c}{a} \\ y \times 2y = \frac{k}{1} \\ 2 {y}^{2} = k \\ 2 {(3)}^{2} = k \\ 2 \times 9 = k \\ 18 = k$

Answer 2

Answer:

this is your answer

Step-by-step explanation:

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