Question

If one root of the quadratic equation (x)square-3x+q=0 is twice the other root, find the value of q.

Answer 1

p(x) = x² - 3x + q = 0
Let α=2β
then,
α+β = -b/a
2β+β = -(-3)/1
3β = 3
β = 1
Now that we have found one root of p(x), we'll substitute in it.
p(x) = x² - 3x + q = 0
⇒ (1)² - 3(1) + q = 0
⇒ 1- 3 + q = 0
⇒ -2 + q = 0
∴ q = 2
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Answer 2

Answer:

p(x) = x² - 3x + q = 0

Let α=2β

then,

α+β = -b/a

2β+β = -(-3)/1

3β = 3

β = 1

Now that we have found one root of p(x), we'll substitute in it.

p(x) = x² - 3x + q = 0

⇒ (1)² - 3(1) + q = 0

⇒ 1- 3 + q = 0

⇒ -2 + q = 0

∴ q = 2

Step-by-step explanation: