Question

If `alpha != beta, alpha^2= 5alpha - 3, beta^2= 5beta-3` then the quation whose roots are `alpha/beta & beta/alpha` is ?

**Ans = `3x^2-19x+3=0`

Answer 1

`3x^2-19x+3=0`

Explanation:

We are effectively told that `alpha` and `beta` are the two roots of the quadratic equation:

`x^2=5x-3`

Rearranged slightly, that is:

`x^2-5x+3 = 0`

So we have:

`0 = x^2-5x+3`

`color(white)(0) = (x-alpha)(x-beta)`

`color(white)(0) = x^2-(alpha+beta)x+alphabeta`

Equating coefficients, that tells us that:

`{ (alpha+beta = 5), (alphabeta = 3) :}`

In addition, note that:

`alpha^2+beta^2 = (5alpha-3)+(5beta-3)`

`color(white)(alpha^2+beta^2) = 5(alpha+beta)-6`

`color(white)(alpha^2+beta^2) = 5(color(blue)(5))-6`

`color(white)(alpha^2+beta^2) = 19`

A quadratic equation with roots `alpha/beta` and `beta/alpha` is:

`(x-alpha/beta)(x-beta/alpha) = 0`

To clear the denominators and give us integer coefficients, we can multiply both sides by `alphabeta` to find:

`0 = alphabeta(x-alpha/beta)(x-beta/alpha)`

`color(white)(0) = (betax-alpha)(alphax-beta)`

`color(white)(0) = alphabetax^2-(alpha^2+beta^2)x+alphabeta`

`color(white)(0) = 3x^2-19x+3`

So we can write a suitable quadratic equation as:

`3x^2-19x+3 = 0`

Answer 2

`3x^2-19x+3=0.`

Explanation:

`alpha^2=5alpha-3.........(1), &, beta^2=5beta-3.................(2)`

`:. (1)-(2) rArr alpha^2-beta^2=5alpha-3-(5beta-3),`

`rArr (alpha-beta)(alpha+beta)=5(alpha-beta),`

Dividing by `(alpha-beta)!=0,...[ because, alpha!=beta],` we get,

`alpha+beta=5...................(star^1).`

Also, `(1)+(2) rArr alpha^2+beta^2=5(alpha+beta)-6,`

`=25-6,............[because, (star^1)]`, i.e.,

`alpha^2+beta^2=(alpha+beta)^2-2alpha*beta=19,`

`:. (star^1) rArr 5^2-2alpha*beta=19,`

`:. alpha*beta=3.........................(star^2).`

Now, `alpha/beta+beta/alpha=(alpha^2+beta^2)/(alpha*beta),`

`=19/3,` and,

`alpha/beta*beta/alpha=1.`

Hence, the Reqd. Quadr. Eqn., having the roots, `alpha/beta, beta/alpha,` is given by,

`x^2-(alpha/beta+beta/alpha)x+(alpha/beta*beta/alpha)=0,` i.e.,

`x^2-(19/3)x+1=0, or, 3x^2-19x+3=0,` as Respected

George C. Sir has already derived.

Enjoy Maths.!