Question

plx help me is solving this
according to 10 board ​

Answer 1

Answer:

g(x)=x²-2mn--(m²-n²)²

Step-by-step explanation:

Let f(x)=2x²+2(m+n)x+m²+n²

p and q are zeros of f(x)

so p+q=-2(m+n)/2=-(m+n)

pq=m²+n²/2

Now the polynomial with  zeros

let h= (p+q)²,and k=(p-q)²

sum of zeros h+k=(p+q)²+(p-q)²=2(p²+q²)

=2 [ (p+q)²-2pq]=(m+n)²-2(m²+n²)/2=(m+n)²-m²-n²=2mn

so h+k=2mn------------------(1)

product of zeros=

hk=(p+q)²*(p-q²

=(p²-q²)²

=(p+q)²(p-q)²

=(p+q)²[ (p+q)²-4pq]

Putting value from above

hk=(m+n)²[ (m+n)²-4(m²+n²)/2)

=(m+n)²[(m+n)²-2m²-2n²]

=(m+n)²(-m²-n²+2mn

=-(m-n)2(m²+n²-2mn)

=-(m-n)2(m-n)2

hk=-(m²-n²)²-----------------------(2)

So the polynomial with h and k as zeros will be

g(x)=x²-(h+k)x+hk

g(x)=x²-2mn--(m²-n²)²