If alpha and beta are roots of the eqⁿ 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.

Tunapanua usemi $(\alpha - 2\beta)(\beta - 2\alpha)$:

$$(\alpha - 2\beta)(\beta - 2\alpha) = \alpha\beta - 2\alpha^2 - 2\beta^2 + 4\alpha\beta$$ $$= 5\alpha\beta - 2(\alpha^2 + \beta^2)$$

Tunajua kuwa $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$. Badilisha hii kwenye usemi:

$$5\alpha\beta - 2((\alpha + \beta)^2 - 2\alpha\beta)$$ $$= 5\alpha\beta - 2(\alpha + \beta)^2 + 4\alpha\beta$$ $$= 9\alpha\beta - 2(\alpha + \beta)^2$$

Sasa tunabadilisha $\alpha + \beta = -\frac{b}{a}$ na $\alpha\beta = \frac{c}{a}$:

$$9\left(\frac{c}{a}\right) - 2\left(-\frac{b}{a}\right)^2$$ $$= \frac{9c}{a} - 2\left(\frac{b^2}{a^2}\right)$$ $$= \frac{9c}{a} - \frac{2b^2}{a^2}$$

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:

$$\boxed{\frac{9ac - 2b^2}{a^2}}$$

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:

$$\boxed{2b^2 = 9ac}$$