Form the quadratic equation whose roots are the squares of the sum of the roots and square of the difference of the root of the equation #2x^2+2(m+n)x+m^2+n^2=0#.?

1 Answer
Jun 1, 2016

Answer:

Desired equation is #x^2-4mnx-(m^2-n^2)^2=0#

Explanation:

Let #alpha# and #beta# be the roots of the equation #2x^2+2(m+n)x+m^2+n^2=0#

As such #alpha+beta=-2(m+n)/2=-(m+n)#

and #alphaxxbeta=(m^2+n^2)/2#

We have to find the equation whose roots are #(alpha+beta)^2# and #(alpha-beta)^2#.

Sum of these roots will be #(alpha+beta)^2+(alpha-beta)^2=2(alpha^2+beta^2)#

= #2((alpha+beta)^2-2alphabeta)=2((-(m+n))^2-2(m^2+n^2)/2)#

= #2(m+n)^2-2(m^2+n^2)=4mn#

Product of these roots will be #(alpha+beta)^2xx(alpha-beta)^2#

= #(alpha^2-beta^2)^2=(alpha^2+beta^2)^2-4alpha^2beta^2#

= #((alpha+beta)^2-2alphabeta)^2-4alpha^2beta^2#

= #((-m-n)^2-2(m^2+n^2)/2)^2-4((m^2+n^2)/2)^2#

= #4m^2n^2-m^4-n^4-2m^2n^2=-m^4-n^4+2m^2n^2#

= #-(m^2-n^2)^2#

Hence equation will be

#x^2-#(sum of roots)#x+#product of roots#=0# or

#x^2-4mnx-(m^2-n^2)^2=0#