If alpha and beta are roots of the eq^n 9x2+bx+c=0, express (alpha - 2beta)(beta - 2alpha) in terms of a, b, c and deduce the condition for the root to be twice that of the other.

Kutokana na mlinganyo wa quadratic ax^2 + bx + c = 0, tunajua yafuatayo kuhusu mizizi na : Jumla ya mizizi: + = -(b)/(a) Zao la mizizi: = (c)/(a) Step 1: Panua usemi uliotolewa. Tunapanua usemi ( - 2)( - 2): ( - 2)( - 2) = - 2^2 - 2^2 + 4 = 5 - 2(^2 + ^2) Step 2: Badilisha ^2 + ^2 kwa kutumia jumla na zao la mizizi. Tunajua kuwa ^2 + ^2 = ( + )^2 - 2. Badilisha hii kwenye usemi: 5 - 2(( + )^2 - 2) = 5 - 2( + )^2 + 4 = 9 - 2( + )^2 Step 3: Badilisha jumla na zao la mizizi kwa kutumia a, b, c. Sasa tunabadilisha + = -(b)/(a) na = (c)/(a): 9((c)/(a)) - 2(-(b)/(a))^2 = (9c)/(a) - 2((b^2)/(a^2)) = (9c)/(a) - (2b^2)/(a^2) Step 4: Rahisisha usemi kwa kupata kigawo cha kawaida. Ili kurahisisha, tunapata kigawo cha kawaida a^2: = (9c · a)/(a · a) - (2b^2)/(a^2) = (9ac - 2b^2)/(a^2) Hivyo, usemi katika a, b, c ni: (9ac - 2b^2)/(a^2) Step 5: Pata sharti kwa mizizi moja kuwa mara mbili ya nyingine. Ikiwa mizizi moja ni mara mbili ya nyingine, kwa mfano, = 2, basi usemi ( - 2)( - 2) utakuwa: (2 - 2)( - 2(2)) = (0)( - 4) = 0 × (-3) = 0 Kwa hiyo, tunalinganisha usemi uliopatikana na sifuri: (9ac - 2b^2)/(a^2) = 0 Kwa kuwa a ≠ 0 (kwa mlinganyo kuwa quadratic), tunaweza kuzidisha pande zote mbili kwa a^2: 9ac - 2b^2 = 0 2b^2 = 9ac Hivyo, sharti ni: 2b^2 = 9ac Tuma swali linalofuata 📸