Whakaoti i te x^2-11x=-28

$$x ^ { 2 } - 11 x = - 28$$

Whakaoti mō x

$$x=4$$

$$x=7$$

Ngā Upane Whakamahi Tauwehe

Me tāpiri te 28 ki ngā taha e rua.

$$x^{2}-11x+28=0$$

Hei whakaoti i te whārite, whakatauwehea te x^{2}-11x+28 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.

$$a+b=-11 ab=28$$

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 28.

$$-1,-28 -2,-14 -4,-7$$

Tātaihia te tapeke mō ia takirua.

$$-1-28=-29 -2-14=-16 -4-7=-11$$

Ko te otinga te takirua ka hoatu i te tapeke -11.

$$a=-7 b=-4$$

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

$$\left(x-7\right)\left(x-4\right)$$

Hei kimi otinga whārite, me whakaoti te x-7=0 me te x-4=0.

$$x=7 x=4$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

Me tāpiri te 28 ki ngā taha e rua.

$$x^{2}-11x+28=0$$

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei x^{2}+ax+bx+28. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-11 ab=1\times 28=28$$

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 28.

$$-1,-28 -2,-14 -4,-7$$

Tātaihia te tapeke mō ia takirua.

$$-1-28=-29 -2-14=-16 -4-7=-11$$

Ko te otinga te takirua ka hoatu i te tapeke -11.

$$a=-7 b=-4$$

Tuhia anō te x^{2}-11x+28 hei \left(x^{2}-7x\right)+\left(-4x+28\right).

$$\left(x^{2}-7x\right)+\left(-4x+28\right)$$

Tauwehea te x i te tuatahi me te -4 i te rōpū tuarua.

$$x\left(x-7\right)-4\left(x-7\right)$$

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

$$\left(x-7\right)\left(x-4\right)$$

Hei kimi otinga whārite, me whakaoti te x-7=0 me te x-4=0.

$$x=7 x=4$$

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.

$$x^{2}-11x=-28$$

Me tāpiri 28 ki ngā taha e rua o te whārite.

$$x^{2}-11x-\left(-28\right)=-28-\left(-28\right)$$

Mā te tango i te -28 i a ia ake anō ka toe ko te 0.

$$x^{2}-11x-\left(-28\right)=0$$

Tango -28 mai i 0.

$$x^{2}-11x+28=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, -11 mō b, me 28 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 28}}{2}$$

Pūrua -11.

$$x=\frac{-\left(-11\right)±\sqrt{121-4\times 28}}{2}$$

Whakareatia -4 ki te 28.

$$x=\frac{-\left(-11\right)±\sqrt{121-112}}{2}$$

Tāpiri 121 ki te -112.

$$x=\frac{-\left(-11\right)±\sqrt{9}}{2}$$

Tuhia te pūtakerua o te 9.

$$x=\frac{-\left(-11\right)±3}{2}$$

Ko te tauaro o -11 ko 11.

$$x=\frac{11±3}{2}$$

Nā, me whakaoti te whārite x=\frac{11±3}{2} ina he tāpiri te ±. Tāpiri 11 ki te 3.

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

Whakawehe 14 ki te 2.

$$x=7$$

Nā, me whakaoti te whārite x=\frac{11±3}{2} ina he tango te ±. Tango 3 mai i 11.

$$x=\frac{8}{2}$$

Whakawehe 8 ki te 2.

$$x=4$$

Kua oti te whārite te whakatau.

$$x=7 x=4$$

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.

$$x^{2}-11x=-28$$

Whakawehea te -11, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{11}{2}. Nā, tāpiria te pūrua o te -\frac{11}{2} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}-11x+\left(-\frac{11}{2}\right)^{2}=-28+\left(-\frac{11}{2}\right)^{2}$$

Pūruatia -\frac{11}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-11x+\frac{121}{4}=-28+\frac{121}{4}$$

Tāpiri -28 ki te \frac{121}{4}.

$$x^{2}-11x+\frac{121}{4}=\frac{9}{4}$$

Tauwehea te x^{2}-11x+\frac{121}{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}.

$$\left(x-\frac{11}{2}\right)^{2}=\frac{9}{4}$$

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

$$\sqrt{\left(x-\frac{11}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Whakarūnātia.

$$x-\frac{11}{2}=\frac{3}{2} x-\frac{11}{2}=-\frac{3}{2}$$

Me tāpiri \frac{11}{2} ki ngā taha e rua o te whārite.

$$x=7 x=4$$