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 $\alpha$ na $\beta$: Jumla ya mizizi: $\alpha + \beta = -\frac{b}{a}$ Zao la mizizi: $\alpha\beta = \frac{c}{a}$

Step 1: Panua usemi uliotolewa. Tunapanua usemi $(\alpha - 2\beta)(\beta - 2\alpha)$: $(\alpha - 2\beta)(\beta - 2\alpha) = \alpha\beta - 2\alpha^2 - 2\beta^2 + 4\alpha\beta$ $= 5\alpha\beta - 2(\alpha^2 + \beta^2)$

Step 2: Badilisha $\alpha^2 + \beta^2$ kwa kutumia jumla na zao la mizizi. Tunajua kuwa $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$. Badilisha hii kwenye usemi: $5\alpha\beta - 2((\alpha + \beta)^2 - 2\alpha\beta)$ $= 5\alpha\beta - 2(\alpha + \beta)^2 + 4\alpha\beta$ $= 9\alpha\beta - 2(\alpha + \beta)^2$

Step 3: Badilisha jumla na zao la mizizi kwa kutumia $a, b, c$. Sasa tunabadilisha $\alpha + \beta = -\frac{b}{a}$ na $\alpha\beta = \frac{c}{a}$: $9\left(\frac{c}{a}\right) - 2\left(-\frac{b}{a}\right)^2$ $= \frac{9c}{a} - 2\left(\frac{b^2}{a^2}\right)$ $= \frac{9c}{a} - \frac{2b^2}{a^2}$

Step 4: Rahisisha usemi kwa kupata kigawo cha kawaida. Ili kurahisisha, tunapata kigawo cha kawaida $a^2$: $= \frac{9c \cdot a}{a \cdot a} - \frac{2b^2}{a^2}$ $= \frac{9ac - 2b^2}{a^2}$ Hivyo, usemi katika $a, b, c$ ni: $\boxed{\frac{9ac - 2b^2}{a^2}}$

Step 5: Pata sharti kwa mizizi moja kuwa mara mbili ya nyingine. Ikiwa mizizi moja ni mara mbili ya nyingine, kwa mfano, $\alpha = 2\beta$, basi usemi $(\alpha - 2\beta)(\beta - 2\alpha)$ utakuwa: $(2\beta - 2\beta)(\beta - 2(2\beta))$ $= (0)(\beta - 4\beta)$ $= 0 \times (-3\beta)$ $= 0$ Kwa hiyo, tunalinganisha usemi uliopatikana na sifuri: $\frac{9ac - 2b^2}{a^2} = 0$ Kwa kuwa $a \neq 0$ (kwa mlinganyo kuwa quadratic), tunaweza kuzidisha pande zote mbili kwa $a^2$: $9ac - 2b^2 = 0$ $2b^2 = 9ac$ Hivyo, sharti ni: $\boxed{2b^2 = 9ac}$ Tuma swali linalofuata 📸