Question

One of the roots of the quadratic equation dx^2+cx+p=0 is twice the other, find the relationship between d, c and p

Answer 1

`c^2 = 9dp`

Explanation:

Given

`dx^2 + cx + p = 0`

Let the roots be
`\alpha` and
`\beta`

So:

`\alpha = 2\beta`

Required

Determine the relationship between d, c and p

`dx^2 + cx + p = 0`

Divide through by d

`(dx^2)/(d) + (cx)/(d) + (p)/(d) = 0`

`x^2 + (c)/(d)x + (p)/(d) = 0`

A quadratic equation has the form:

`x^2 - (\alpha + \beta)x + \alpha \beta = 0`

So:

`x^2 - (2\beta+ \beta)x + \beta*\beta = 0`

`x^2 - (3\beta)x + \beta^2 = 0`

So, we have:

`(c)/(d) = -3\beta` -- (1)

and

`(p)/(d) = \beta^2` -- (2)

Make
`\beta` the subject in (1)

`(c)/(d) = -3\beta`

`\beta = -(c)/(3d)`

Substitute
`\beta = -(c)/(3d)` in (2)

`(p)/(d) = (-(c)/(3d))^2`

`(p)/(d) = (c^2)/(9d^2)`

Multiply both sides by d

`d * (p)/(d) = (c^2)/(9d^2)*d`

`p = (c^2)/(9d)`

Cross Multiply

`9dp = c^2`

or

`c^2 = 9dp`

Hence, the relationship between d, c and p is:
`c^2 = 9dp`