Question

Is there a simplest way to solve this ?

Given `alpha^2` + `beta^2` = 17 and `alphabeta` = 4 , where `alpha` and `beta` are positive values. Form the quadratic equation which has the roots `alpha/2` and `beta/2`

Answer 1

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

Explanation:

Suppose that, `S=alpha/2+beta/2..............(1),` and,

`P=alpha/2*beta/2.........................................(2).`

Then, `x^2-Sx+Q=0............(3),` will be the reqd. quadr. eqn.

`(1) rArr alpha+beta=2S.`

Given that, `alpha^2+beta^2=17, and, alpha*beta=4`

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

`:. 17=(2S)^2-2*4=4S^2-8 rArr 4S^2=17+8=25,`

`:. S=pm5/2...........................(4).`

But, given that, `alpha >0, and, beta >0.`

`:. S=+5/2..............................(4').`

Clearly, `P=1/4*alpha*beta=1/4*4=1...............(4'').`

Therefore, by `(3). (4), and, (4''),` we have,

`x^2-5/2*x+1=0, or, 2x^2-5x+2=0,` as the desired eqn.

Enjoy Maths.!

Answer 2

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

Explanation:

As an Aliter, consider the following Solution :

Given that, `alpha^2+beta^2=17......(1), and, alpha*beta=4.......(2).`

`(2) rArr beta=4/alpha...............(2').`

`(1), and, (2') rArr alpha^2+16/alpha^2=17.`

`:. alpha^4+16=17alpha^2, i.e., alpha^4-17alpha^2+16=0.`

`:.(alpha^2-16)(alpha^2-1)=0.`

`;. alpha^2=16, or, alpha^2=1.`

Knowing that, `alpha > 0," we have, "alpha=4, or, 1.`

By`(2')," then, "beta=1, or, beta=4.`

`:. (alpha,beta)=(4,1), or, (1,4).`

`:.," in either case, "alpha/2+beta/2=5/2, and, alpha/2*beta/2=1,`

giving the desired quadr. eqn.,

`x^2-5/2*x+1=0, or, 2x^2-5x+2=0.`

Enjoy Maths.!

Answer 3

See below.

Explanation:

Another approach

`4((alpha+beta)/2)^2 = alpha^2+beta^2+2alpha beta` then

`(alpha+beta)/2=1/2 sqrt(alpha^2+beta^2+2alpha beta) = 5/2`

Now if `(y-y_1)(y-y_2) = y^2-(y_1+y_2)y+y_1y_2=0` we substitute

`y_1+y_2 -> 5/2` and `y_1y_2 = (alpha beta)/4 = 1` and we get

`y^2-5/2y+1=0` which is the sought polynomial.

Answer 4

Monic quadratic equation:

`x^2-5/2x+1 = 0`

or integer coefficients:

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

Explanation:

Another formulation:

Given:

`{ (alpha^2+beta^2 = 17), (alphabeta=4), (alpha > 0), (beta > 0) :}`

Note that since `alpha, beta > 0` we have:

`alpha+beta = sqrt((alpha+beta)^2)`

So:

`(x-alpha/2)(x-beta/2) = x^2-(alpha+beta)/2x+(alphabeta)/4`

`color(white)((x-alpha/2)(x-beta/2)) = x^2-(sqrt((alpha+beta)^2))/2x+(alphabeta)/4`

`color(white)((x-alpha/2)(x-beta/2)) = x^2-sqrt(alpha^2+beta^2+2alphabeta)/2x+(alphabeta)/4`

`color(white)((x-alpha/2)(x-beta/2)) = x^2-sqrt(17+2(4))/2x+4/4`

`color(white)((x-alpha/2)(x-beta/2)) = x^2-sqrt(25)/2x+1`

`color(white)((x-alpha/2)(x-beta/2)) = x^2-5/2x+1`

This polynomial has the required zeros, so we can write:

`x^2-5/2x+1 = 0`

or multiply by `2` to get a polynomial with integer coefficients:

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