Tolong bantu jawab no 3 4 5 6 7

Tolong bantu jawab no 3 4 5 6 7

Jawaban Terkonfirmasi

Nomor 3.
Komposisi:
$x_1+x_2=,^5/_3 \ x_1x_2=,^6/_3=2$
Maka,
a.
$$begin{align}left[x-frac{3}{x_1}right]left[x-frac{3}{x_2}right]&=0 \ x^2-left[frac{3}{x_1}+frac{3}{x_2}right]x+frac{9}{x_1x_2}&=0 \ x^2-frac{3(x_1+x_2)}{x_1x_2}x+frac{9}{x_1x_2}&=0 \ x_1x_2x^2-3(x_1+x_2)x+9&=0 \ 2x^2-3(frac{5}{3})x+9&=0 \ 2x^2-5x+9&=0 end{align}$

b. (Mungkin soal dan jawaban sedikit aneh)
Asumsi $x_1>x_2$
Sehingga, komposisi baru:
$x_1-x_2=,^{sqrt{D}}/_a \ x_1-x_2=, ^{sqrt{-47}}/_3$
Didapat:
$x_1^2-x_2^2=(x_1+x_2)(x_1-x_2) \ =,^5/_3timessqrt{-47}$
Sehingga,
$$begin{align}(x-(x_1x_2))(x-(x_1^2-x_2^2))&=0 \ x^2-(x_1x_2+(x_1^2-x_2^2))x+(x_1x_2)(x_1^2-x_2)^2&=0 \ x^2-(2+frac{5}{3}sqrt{-47})x+frac{10}{3}sqrt{-47}&=0 \ 3x^2-(6+5sqrt{-47})x+10sqrt{-47}&=0 end{align}$

Nomor 4.
Koordinat puncak diberikan:
$$begin{align}text{Puncak : }&left[-frac{b}{2a},-frac{b^2-4ac}{2a}right] \ &left[-frac{6}{2(4)},-frac{6^2-4(4)(-5)}{2(4)}right] \ &left[-frac{6}{8},-frac{36+80}{8}right] \ &left[-frac{3}{4},-frac{116}{8}right] \ &left[-frac{3}{4},-frac{29}{2}right] end{align}$

Nomor 5.
Diberikan demikian:
$$begin{align}x_1-x_2&=5 \ pmfrac{sqrt{D}}{a}&=5 \ pmfrac{sqrt{D}}{1}&=5 \ pmsqrt{b^2-4ac}&=5 \ b^2-4ac&=5^2 \ b^2-4ac&=25 \ m^2-4(1)(14)&=25 \ m^2-56&=25 \ m^2-81&=0 \ (m+9)(m-9)&=0end{align}$
Sehingga,
m = -9
m = 9

Nomor 6.
$x_1^2+x_2^2=(x_1+x_2)^2-2x_1x_2$
Komposisi:
$x_1+x_2=, ^k/_3 \\ x_1x_2=,^{k+4}/_3$
Sehingga,
$$begin{align}x_1^2+x_2^2&=(x_1+x_2)^2-2x_1x_2 \ &=left[frac{k}{3}right]^2-2timesfrac{k+4}{3} \ &=frac{k^2}{9}-frac{2k+8}{3} \ &=frac{1}{9}(k^2-6k-24) \ &=frac{1}{9}k^2-frac{2}{3}k-frac{8}{3}end{align}$
Dengan nilai minimum:
$$begin{align}text{Minimum : }&-frac{b^2-4ac}{4a} \ &-frac{(-frac{2}{3})^2-4(frac{1}{9})(-frac{8}{3})}{2(frac{1}{9})} \ &-frac{frac{4}{9}+frac{32}{27}}{frac{2}{9}} \ &-frac{4+frac{32}{3}}{2} \ &-frac{frac{12+32}{3}}{2} \ &-frac{44}{6} \ text{Minimum : }&-frac{22}{3}end{align}$

Nomor 7.
Dengan properti:
2 akar dan melalui satu titik:
$y=a(x-x_1)(x-x_2)$
Akarnya adalah -3, dan 6
Sehingga,
$y=a(x-(-3))(x-6) \ y=a(x+3)(x-6)$
Substitusikan nilai x dan y dari titik lain:
$-18=a(0+3)(0-6) \ -18=-18a \ a=1$
Sehingga,
$y=a(x+3)(x-6) \ y=1(x+3)(x-6) \ boxed{y=x^2-3x-18}$

Bonus, Nomor 8.
Rumus:
$boxed{y=a(x-x_p)^2+y_p}$
Maka,
$y=a(x-3)^2+5$
Substitusikan titik lainnya:
$14=a(0-3)^2+5 \ 14-5=a(3)^2 \ 9a=9 \ a=1$
Sehingga, persamaannya:
$y=a(x-3)^2+5 \ y=1(x-3)^2+5 \ y=x^2-6x+9+5 \ boxed{y=x^2-6x+14}$