Question

Find the condition that one root of px^2-qx-r=0, (p is not equal to 0) maybe double of the other?

Answer 1

`2q^2=-9rp`

Explanation:

`px^2-qx-r=0`

given one root is double the otehr

# "ie roots are " alpha, 2alpha#

`alpha +2alpha=3alpha= - -q/p=q/p ----(1)`

`alpha xx 2alpha =2alpha ^2=-r/p ---(2)`

from `(1)`

`alpha =q/(3p)`

substitute into `(2)`

`2(q/(3p))^2=-r/p`

`(2q^2)/(9p^2)=-r/p`

`=> 2q^2=-9rp`

Answer 2

`r=(-2q^2)/(9p)`

Explanation:

Calling one root `b` and the other `2b`:

`a(x-b)(x-2b)=px^2-qx-r`

`a(x^2-3bx+2b^2)=px^2-qx-r`

`ax^2-3abx+2ab^2=px^2-qx-r`

Comparing coefficients, we see that

`{(a=p),(-3ab=-q),(-r=2ab^2):}`

From the second equation we see that `b=q/(3a)=q/(3p)`.

Then, `r=-2ab^2=-2p(q/(3p))^2=(-2q^2)/(9p)`