Whakaoti i te x^2-13x+42=0

$$x ^ { 2 } - 13 x + 42 = 0$$

Whakaoti mō x

$$x=6$$

$$x=7$$

Ngā Upane Whakamahi Tauwehe

Hei whakaoti i te whārite, whakatauwehea te x^{2}-13x+42 mā te whakamahi i te tātai x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-13 ab=42$$

I te mea kua tōrunga te ab, he ōrite te tohu o a me b. I te mea kua tōraro te a+b, he tōraro hoki a a me b. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua 42.

$$-1,-42 -2,-21 -3,-14 -6,-7$$

Tātaihia te tapeke mō ia takirua.

$$-1-42=-43 -2-21=-23 -3-14=-17 -6-7=-13$$

Ko te otinga te takirua ka hoatu i te tapeke -13.

$$a=-7 b=-6$$

Me tuhi anō te kīanga whakatauwehe \left(x+a\right)\left(x+b\right) mā ngā uara i tātaihia.

$$\left(x-7\right)\left(x-6\right)$$

Hei kimi otinga whārite, me whakaoti te x-7=0 me te x-6=0.

$$x=7 x=6$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei x^{2}+ax+bx+42. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-13 ab=1\times 42=42$$

I te mea kua tōrunga te ab, he ōrite te tohu o a me b. I te mea kua tōraro te a+b, he tōraro hoki a a me b. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua 42.

$$-1,-42 -2,-21 -3,-14 -6,-7$$

Tātaihia te tapeke mō ia takirua.

$$-1-42=-43 -2-21=-23 -3-14=-17 -6-7=-13$$

Ko te otinga te takirua ka hoatu i te tapeke -13.

$$a=-7 b=-6$$

Tuhia anō te x^{2}-13x+42 hei \left(x^{2}-7x\right)+\left(-6x+42\right).

$$\left(x^{2}-7x\right)+\left(-6x+42\right)$$

Tauwehea te x i te tuatahi me te -6 i te rōpū tuarua.

$$x\left(x-7\right)-6\left(x-7\right)$$

Whakatauwehea atu te kīanga pātahi x-7 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(x-7\right)\left(x-6\right)$$

Hei kimi otinga whārite, me whakaoti te x-7=0 me te x-6=0.

$$x=7 x=6$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Ko ngā whārite katoa o te āhua ax^{2}+bx+c=0 ka taea te whakaoti mā te whakamahi i te tikanga tātai pūrua: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. E rua ngā otinga ka puta i te tikanga tātai pūrua, ko tētahi ina he tāpiri a ±, ā, ko tētahi ina he tango.

$$x^{2}-13x+42=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, -13 mō b, me 42 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\times 42}}{2}$$

Pūrua -13.

$$x=\frac{-\left(-13\right)±\sqrt{169-4\times 42}}{2}$$

Whakareatia -4 ki te 42.

$$x=\frac{-\left(-13\right)±\sqrt{169-168}}{2}$$

Tāpiri 169 ki te -168.

$$x=\frac{-\left(-13\right)±\sqrt{1}}{2}$$

Tuhia te pūtakerua o te 1.

$$x=\frac{-\left(-13\right)±1}{2}$$

Ko te tauaro o -13 ko 13.

$$x=\frac{13±1}{2}$$

Nā, me whakaoti te whārite x=\frac{13±1}{2} ina he tāpiri te ±. Tāpiri 13 ki te 1.

$$x=\frac{14}{2}$$

Whakawehe 14 ki te 2.

$$x=7$$

Nā, me whakaoti te whārite x=\frac{13±1}{2} ina he tango te ±. Tango 1 mai i 13.

$$x=\frac{12}{2}$$

Whakawehe 12 ki te 2.

$$x=6$$

Kua oti te whārite te whakatau.

$$x=7 x=6$$

Ngā Upane mō te Whakaoti i te Pūrua

Ko ngā whārite pūrua pēnei i tēnei nā ka taea te whakaoti mā te whakaoti i te pūrua. Hei whakaoti i te pūrua, ko te whārite me mātua tuhi ki te āhua x^{2}+bx=c.

$$x^{2}-13x+42=0$$

Me tango 42 mai i ngā taha e rua o te whārite.

$$x^{2}-13x+42-42=-42$$

Mā te tango i te 42 i a ia ake anō ka toe ko te 0.

$$x^{2}-13x=-42$$

Whakawehea te -13, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{13}{2}. Nā, tāpiria te pūrua o te -\frac{13}{2} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}-13x+\left(-\frac{13}{2}\right)^{2}=-42+\left(-\frac{13}{2}\right)^{2}$$

Pūruatia -\frac{13}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-13x+\frac{169}{4}=-42+\frac{169}{4}$$

Tāpiri -42 ki te \frac{169}{4}.

$$x^{2}-13x+\frac{169}{4}=\frac{1}{4}$$

Tauwehea te x^{2}-13x+\frac{169}{4}. Ko te tikanga, ina ko x^{2}+bx+c he pūrua tika, ka taea te tauwehe i ngā wā katoa hei \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{13}{2}\right)^{2}=\frac{1}{4}$$

Tuhia te pūtakerua o ngā taha e rua o te whārite.

$$\sqrt{\left(x-\frac{13}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Whakarūnātia.

$$x-\frac{13}{2}=\frac{1}{2} x-\frac{13}{2}=-\frac{1}{2}$$

Me tāpiri \frac{13}{2} ki ngā taha e rua o te whārite.

$$x=7 x=6$$