Roots of x2−2cx+ab=0 are real and equal ⇒ Discriminant >0 ⇒‌‌(−2c)2−4(1)(ab)>0 ⇒4c2−4ab>0⇒c2>ab>0 Now, discriminant of x2−2(a+b)‌x+a2+b2+2c2=0 D‌‌=[−2(a+b)]2−4(1)(a2+b2+2c2) ‌‌=4(a+b)2−4(a2+b2+2c2) ‌‌=8ab−8c2 ‌‌=−8(c2−ab)<0[‌ as ‌c2−ab>0] ⇒‌ Roots of ‌x2−2(a+b)x+a2+b2+2c2‌ is ‌