Question

If the roots of `2x^2-3x+6 = 0` are `p` and `q`, then what quadratic equation has roots `p^2+2` and `q^2+2` ?

Answer 1

`4x^2-49x+118=0`

Explanation:

.

`2x^2-3x-6=0`

`x=(3+-sqrt(9-4(2)(-6)))/4=(3+-sqrt(9+48))/4=(3+-sqrt57)/4`

`p=(3+sqrt57)/4`

`q=(3-sqrt57)/4`

`p^2+2=((3+sqrt57)/4)^2+2=(9+57+6sqrt57)/16+2`

`p^2+2=(66+6sqrt57)/16+2=(33+3sqrt57)/8+2=(33+3sqrt57+16)/8`

`p^2+2=(49+3sqrt57)/8`

`q^2+2=((3-sqrt57)/4)^2+2=(9+57-6sqrt57)/16+2`

`q^2+2=(33-3sqrt57)/8+2=(33-3sqrt57+16)/8`

`q^2+2=(49-3sqrt57)/8`

The equation would be:

`(x-(49+3sqrt57)/8)(x-(49-3sqrt57)/8)=0`

`x^2-x((49-3sqrt57)/8)-x((49+3sqrt57)/8)+(49^2-513)/64=0`

`x^2-x((49-3sqrt57+49+3sqrt57)/8)+59/2=0`

`x^2-98/8x+59/2=0`

`x^2-49/4x+59/2=0`

`4x^2-49x+118=0`

Answer 2

`4x^2-49x+118 = 0`

Explanation:

Dividing through by `2` we get:

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

`color(white)(x^2-3/2x-3) = x^2-(p+q)x+pq`

Hence:

`{ (p+q = 3/2), (pq = -3) :}`

Then:

`p^2+q^2 = (p+q)^2-2pq = (3/2)^2-2(-3) = 9/4+6 = 33/4`

So:

`(p^2+2) + (q^2+2) = 33/4+4 = 49/4`

`(p^2+2)(q^2+2) = p^2q^2+2(p^2+q^2)+4`

`color(white)((p^2+2)(q^2+2)) = (-3)^2+2(33/4)+4 = 9+33/2+4 = 59/2`

Hence a quadratic equation with roots `p^2+2` and `q^2+2` is:

`x^2-49/4x+59/2 = 0`

Multiplying through by `4` we get:

`4x^2-49x+118 = 0`