Question

If the root of the quadratic equation x^2 + 2px +mn= 0 are real and equal. show that the roots of the quadratic equation x^2 -2(m+n)x + (m^2 +n^2 + 2p^2 )= 0 are equal roots...answer plzz

Answer 1

That is ur perfect solution....

In the first equation we discriminate it
As by b^2-4ac=0
On putting the values of first equation in this we get a relation
P^2=MN---------------------(1)
Now in second equation which we have to prove equal roots, so for proving it we have to prove it's discrimination of b^2-4ac equal to zero then it will be proved that this has equal roots,,, so now solving,,,
On putting the values of first equation in b^2-4ac.....that
{-2(m+n)}^2 - 4x1x(m^2+n^2+2p^2)

4(m+n)^2 - 4(m^2+n^2+2p^2)
Taking 4 common

{ (m+n)^2 - (m^2+n^2+2p^2) }
Now we know the identity (a+b) ^2 and we had proved above that p^=में so we put both in this

4[ m^2 + n^2 + 2mn - m^2 - n^2 - 2mn]

In the bracket like terms cancel each other and we get
4(0)=0
0=0
Lhs=rhs

Hence on discriminating we get it equal to zero this shows that this equation has equal roots.. ... Hence prooved

OK miss mixture,, that is ur perfect answer and if u dont understand until u can again ask ur query in comments,,, :-)

All the best for ur exam