Question

Kuis (medium)
Dik: Rx² - Sx + T = 0
Ditanya:

[tex]\displaystyle\rm(a.)\:\:\:\frac{1}{x_1}+\frac{1}{x_2}\:\:=\:\:.........\\\\(b.)\:\:\:\frac{x_1}{(x_2)^2}+\frac{x_2}{(x_1)^2}\:\:=\:\:.........\\\\(c.)\:\:\:(x_1-x_2)^2\:\:=\:\:.........[/tex]

Answer 1

Jawaban:

[tex]a.) \: \frac{1}{x_1}+\frac{1}{x_2}[/tex]

[tex] = \dfrac{S}{T} [/tex]

[tex]b.)\:\frac{x_1}{(x_2)^2}+\frac{x_2}{(x_1)^2}[/tex]

[tex] = \dfrac{2T}{R} [/tex]

[tex]c.) \: (x_1-x_2)^2[/tex]

[tex] = \dfrac{{S}^{2}}{ {R}^{2} } - \dfrac{4T}{R} [/tex]

Penjelasan dengan langkah-langkah:

Rx² - Sx + T = 0

a = R, b = -S, c = T

Jadi:

Diskriminan (D) = b² - 4ac

= S² - 4RT

[tex]x_1+x_2 = - \frac{b}{a} = \frac{S}{R} [/tex]

[tex]x_1x_2 = \frac{c}{a} = \frac{T}{R} [/tex]

[tex]x_1 - x_2 = \frac{ \sqrt{d} }{a} [/tex]

[tex] = \frac{ \sqrt{ {S}^{2} - 4RT } }{R} [/tex]

Soal:

[tex]a.) \: \frac{1}{x_1}+\frac{1}{x_2}[/tex]

[tex] = \frac{x_2+x_1} {x_1x_2}\to \frac{x_1+x_2}{x_1x_2}[/tex]

[tex] = \dfrac{ \dfrac{S}{R} }{ \dfrac{T}{R} } = \dfrac{S}{R} \times \dfrac{R}{T} [/tex]

[tex] = \dfrac{SR}{RT }[/tex]

[tex] = \dfrac{S}{T} [/tex]

[tex]b.)\:\frac{x_1}{(x_2)^2}+\frac{x_2}{(x_1)^2}[/tex]

[tex] = \dfrac{{x_1}^{3} {x_2}^{3} + {x_2}^{3} {x_1}^{3} }{(x_2)^2(x_1)^2}[/tex]

[tex] = \dfrac{2{(x_1x_2)}^{3} }{ {(x_1x_2)}^{2} } [/tex]

[tex] = \dfrac{2{( \frac{T}{R} )}^{3} }{ {( \frac{T}{R} )}^{2} } [/tex]

[tex] = \dfrac{2 {T}^{3} }{ {R}^{3} } \times \dfrac{ {R}^{2} }{ {T}^{2} } [/tex]

[tex] = \dfrac{2 {T}^{3} {R}^{2} }{ {R}^{3} {T}^{2} } [/tex]

[tex] = \dfrac{2T}{R} [/tex]

[tex]c.) \: (x_1-x_2)^2[/tex]

[tex] = (\frac{ \sqrt{ {S}^{2} - 4RT} }{R} )^2[/tex]

[tex] = \frac{ {(\sqrt{ {S}^{2} - 4RT })}^{2} }{{R}^{2} }[/tex]

[tex] = \dfrac{{S}^{2} - 4RT}{ {R}^{2} } [/tex]

[tex] = \dfrac{{S}^{2}}{ {R}^{2} } - \dfrac{4RT}{ {R}^{2} } [/tex]

[tex] = \dfrac{{S}^{2}}{ {R}^{2} } - \dfrac{4T}{R} [/tex]