If α and β be the roots of the equation 2x2+2(a+b)x+a2+b2=0 then the equation whose roots are (α+β)2 and (α-β)2 is

Sum of roots α+β=-(a+b) and αβ=a2+b22⇒(α+β)2=(a+b)2 and (α-β)2=α2+β2-2αβ 2ab-(a2+b2)=-(a-b)2 Now the required equation whose roots are(α+β)2 and (α-β)2x2={(α+β)2+(α-β)2}x+(α+β)2(α-β)2=0 ⇒x2-{(a+b)2-(a-b)2}x-(a+b)2(a-b)2=0  ⇒x2-4abx-(a2-b2)2=0