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#

4 Answers
Ratnaker Mehta
Aug 27, 2017

# 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.!

Ratnaker Mehta
Aug 27, 2017

# 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.!

Cesareo R.
Aug 27, 2017

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.

George C.
Aug 27, 2017

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#

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