Question

Given that one root is 3 times the others for the quadratic equation `3x^2-2x+p=0`, find (a) the value of `p` and (b) the two roots ?

Answer 1

`p=1/4, x=1/6, 1/2`

Explanation:

Let the roots be `r, 3r`:

`a(x-r)(x-3r)=0`

`ax^2-4arx+3ar^2=0`

`3x^2-2x+p=ax^2-4arx+3ar^2`

`a=3`

`3x^2-2x+p=3x^2-12rx+9r^2`

`12r=2`

`r=1/6=>` `3r=1/2`

`p=9*1/(36)=1/4`

Check:

`3x^2-2x+1/4=0`
`3(x^2-(2/3)x)=-1/4`
`3(x^2(-2/3)x+(1/3)^2)=1/3-1/4=1/12`
`(x-1/3)^2=1/36`
`x-1/3=+-1/6`
`x=1/3+-1/6`
`x=1/3-1/6=2/6-1/6=1/6`
`x=1/3+1/6=2/6+1/6=3/6=1/2`