Given, ‌‌2|x|2+51=|1+20x| Then, following two possible cases arise: Case I When 1+20x>0 ⇒‌‌x>−‌
1
20
∴‌‌2x2+51=(1+20x) ⇒‌‌2x2−20x+50=0 ⇒‌‌x2−10x+25=0‌‌[∵2≠0] ⇒‌‌(x−5)2=0⇒x=5,5 Case II When 1+20x<0 ⇒‌‌x<−‌
1
20
∴‌2x2+51=−(1+20x) ⇒‌‌2x2−20x+52=0 ⇒‌‌x2−10x+26=0 Here, ‌‌D<0 Thus, roots are imaginary. Hence, sum of real roots =5+5=10