C-03-Q064mediumsingle_mcqIf a+b=7a+b = \sqrt{7}a+b=7 and a−b=5a-b = \sqrt{5}a−b=5, 8ab(a2+b2)=?8ab(a^2+b^2) = ?8ab(a2+b2)=?a242424b121212c484848d353535ব্যাখ্যাHere a2+b2=(a+b)2+(a−b)22=7+52=6a^2+b^2=\frac{(a+b)^2+(a-b)^2}{2}=\frac{7+5}{2}=6a2+b2=2(a+b)2+(a−b)2=27+5=6 and 4ab=(a+b)2−(a−b)2=7−5=24ab=(a+b)^2-(a-b)^2=7-5=24ab=(a+b)2−(a−b)2=7−5=2, so ab=12ab=\tfrac12ab=21. Therefore 8ab(a2+b2)=8⋅12⋅6=248ab(a^2+b^2)=8\cdot\tfrac12\cdot6=248ab(a2+b2)=8⋅21⋅6=24.