Question

If `alpha` and `beta` be the value of `x` obtained from the equation `m^2(x^2-x)+2mx+3=0` and if `m_1` and m2 be the two value of m for which alpha and beta are connected by the relation alpha by beta +beta by alpha =4/3 find the value of m1^2/m2+m2^2/m1?

Answer 1

`-68/3`

Explanation:

The equation is

`m^2x^2+(2m-m^2)x+3=0`

This means that for the two roots `alpha` and `beta` we have

`alpha+beta = (m^2-2m)/m^2 qquad"and"qquad alpha beta = 3/m^2`

Thus

`alpha/beta + beta/alpha = (alpha^2+beta^2)/(alpha beta)`
`qquadqquad = ((alpha+beta)^2-2alpha beta)/(alpha beta)`
`qquad qquad = (alpha+beta)^2/(alpha beta)-2`
`qquadqquad = (m^2-2m)^2/m^4 times m^2/3-2`
`qquadqquad = (m^2-2m)^2/(3m^2)-2`
`qquad qquad = (m-2)^2/3-2`

(note that `m ne 0` as is obvious from the original equation, which leads to a contradiction for `m=0`)

and so

`alpha/beta + beta/alpha = 4/3 implies`

`(m-2)^2/3-2=4/3 implies`

`(m-2)^2 = 10 implies`

`m^2-4m-6=0`

(note that we can easily find the two values `m_1` and `m_2` to be `2 pm sqrt 10`, respectively - but the algebra is easier if we proceed by using properties of the quadratic equation)

From this quadratic we have

`m_1+m_2 = 4 qquad"and"qquad m_1m_2=-6`

Hence

`m_1^2/m_2++m_2^2/m_1 = (m_1^3+m_2^3)/(m_1m_2)`
`qquad qquad = ((m_1+m_2)^3-3m_1m_2(m_1+m_2))/(m_1m_2)`
`qquadqquad = (m_1+m_2)^3/(m_1m_2)-3(m_1+m_2)`
`qquadqquad = 4^3/(-6)-3times 4 = -64/6-12 = -68/3`