Whakaoti i te (x-3)^2=(2x+1)^2

$$( x - 3 ) ^ { 2 } = ( 2 x + 1 ) ^ { 2 }$$

Whakaoti mō x

$$x=\frac{2}{3}\approx 0.666666667$$

$$x=-4$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(x-3\right)^{2}.

$$x^{2}-6x+9=\left(2x+1\right)^{2}$$

Whakamahia te ture huarua \left(a+b\right)^{2}=a^{2}+2ab+b^{2} hei whakaroha \left(2x+1\right)^{2}.

$$x^{2}-6x+9=4x^{2}+4x+1$$

Tangohia te 4x^{2} mai i ngā taha e rua.

$$x^{2}-6x+9-4x^{2}=4x+1$$

Pahekotia te x^{2} me -4x^{2}, ka -3x^{2}.

$$-3x^{2}-6x+9=4x+1$$

Tangohia te 4x mai i ngā taha e rua.

$$-3x^{2}-6x+9-4x=1$$

Pahekotia te -6x me -4x, ka -10x.

$$-3x^{2}-10x+9=1$$

Tangohia te 1 mai i ngā taha e rua.

$$-3x^{2}-10x+9-1=0$$

Tangohia te 1 i te 9, ka 8.

$$-3x^{2}-10x+8=0$$

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei -3x^{2}+ax+bx+8. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-10 ab=-3\times 8=-24$$

I te mea kua tōraro te ab, he tauaro ngā tohu o a me b. I te mea kua tōraro te a+b, he nui ake te uara pū o te tau tōraro i tō te tōrunga. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua -24.

$$1,-24 2,-12 3,-8 4,-6$$

Tātaihia te tapeke mō ia takirua.

$$1-24=-23 2-12=-10 3-8=-5 4-6=-2$$

Ko te otinga te takirua ka hoatu i te tapeke -10.

$$a=2 b=-12$$

Tuhia anō te -3x^{2}-10x+8 hei \left(-3x^{2}+2x\right)+\left(-12x+8\right).

$$\left(-3x^{2}+2x\right)+\left(-12x+8\right)$$

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

$$-x\left(3x-2\right)-4\left(3x-2\right)$$

Whakatauwehea atu te kīanga pātahi 3x-2 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(3x-2\right)\left(-x-4\right)$$

Hei kimi otinga whārite, me whakaoti te 3x-2=0 me te -x-4=0.

$$x=\frac{2}{3} x=-4$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(x-3\right)^{2}.

$$x^{2}-6x+9=\left(2x+1\right)^{2}$$

Whakamahia te ture huarua \left(a+b\right)^{2}=a^{2}+2ab+b^{2} hei whakaroha \left(2x+1\right)^{2}.

$$x^{2}-6x+9=4x^{2}+4x+1$$

Tangohia te 4x^{2} mai i ngā taha e rua.

$$x^{2}-6x+9-4x^{2}=4x+1$$

Pahekotia te x^{2} me -4x^{2}, ka -3x^{2}.

$$-3x^{2}-6x+9=4x+1$$

Tangohia te 4x mai i ngā taha e rua.

$$-3x^{2}-6x+9-4x=1$$

Pahekotia te -6x me -4x, ka -10x.

$$-3x^{2}-10x+9=1$$

Tangohia te 1 mai i ngā taha e rua.

$$-3x^{2}-10x+9-1=0$$

Tangohia te 1 i te 9, ka 8.

$$-3x^{2}-10x+8=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi -3 mō a, -10 mō b, me 8 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\left(-3\right)\times 8}}{2\left(-3\right)}$$

Pūrua -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\left(-3\right)\times 8}}{2\left(-3\right)}$$

Whakareatia -4 ki te -3.

$$x=\frac{-\left(-10\right)±\sqrt{100+12\times 8}}{2\left(-3\right)}$$

Whakareatia 12 ki te 8.

$$x=\frac{-\left(-10\right)±\sqrt{100+96}}{2\left(-3\right)}$$

Tāpiri 100 ki te 96.

$$x=\frac{-\left(-10\right)±\sqrt{196}}{2\left(-3\right)}$$

Tuhia te pūtakerua o te 196.

$$x=\frac{-\left(-10\right)±14}{2\left(-3\right)}$$

Ko te tauaro o -10 ko 10.

$$x=\frac{10±14}{2\left(-3\right)}$$

Whakareatia 2 ki te -3.

$$x=\frac{10±14}{-6}$$

Nā, me whakaoti te whārite x=\frac{10±14}{-6} ina he tāpiri te ±. Tāpiri 10 ki te 14.

$$x=\frac{24}{-6}$$

Whakawehe 24 ki te -6.

$$x=-4$$

Nā, me whakaoti te whārite x=\frac{10±14}{-6} ina he tango te ±. Tango 14 mai i 10.

$$x=-\frac{4}{-6}$$

Whakahekea te hautanga \frac{-4}{-6} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

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

Kua oti te whārite te whakatau.

$$x=-4 x=\frac{2}{3}$$

Ngā Upane mō te Whakaoti i te Pūrua

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(x-3\right)^{2}.

$$x^{2}-6x+9=\left(2x+1\right)^{2}$$

Whakamahia te ture huarua \left(a+b\right)^{2}=a^{2}+2ab+b^{2} hei whakaroha \left(2x+1\right)^{2}.

$$x^{2}-6x+9=4x^{2}+4x+1$$

Tangohia te 4x^{2} mai i ngā taha e rua.

$$x^{2}-6x+9-4x^{2}=4x+1$$

Pahekotia te x^{2} me -4x^{2}, ka -3x^{2}.

$$-3x^{2}-6x+9=4x+1$$

Tangohia te 4x mai i ngā taha e rua.

$$-3x^{2}-6x+9-4x=1$$

Pahekotia te -6x me -4x, ka -10x.

$$-3x^{2}-10x+9=1$$

Tangohia te 9 mai i ngā taha e rua.

$$-3x^{2}-10x=1-9$$

Tangohia te 9 i te 1, ka -8.

$$-3x^{2}-10x=-8$$

Whakawehea ngā taha e rua ki te -3.

$$\frac{-3x^{2}-10x}{-3}=-\frac{8}{-3}$$

Mā te whakawehe ki te -3 ka wetekia te whakareanga ki te -3.

$$x^{2}+\left(-\frac{10}{-3}\right)x=-\frac{8}{-3}$$

Whakawehe -10 ki te -3.

$$x^{2}+\frac{10}{3}x=-\frac{8}{-3}$$

Whakawehe -8 ki te -3.

$$x^{2}+\frac{10}{3}x=\frac{8}{3}$$

Whakawehea te \frac{10}{3}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{5}{3}. Nā, tāpiria te pūrua o te \frac{5}{3} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}+\frac{10}{3}x+\left(\frac{5}{3}\right)^{2}=\frac{8}{3}+\left(\frac{5}{3}\right)^{2}$$

Pūruatia \frac{5}{3} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}+\frac{10}{3}x+\frac{25}{9}=\frac{8}{3}+\frac{25}{9}$$

Tāpiri \frac{8}{3} ki te \frac{25}{9} mā te kimi i te tauraro pātahi me te tāpiri i ngā taurunga. Ka whakaiti i te hautanga ki ngā kīanga tau iti rawa e taea ana.

$$x^{2}+\frac{10}{3}x+\frac{25}{9}=\frac{49}{9}$$

Tauwehea x^{2}+\frac{10}{3}x+\frac{25}{9}. Ko te tikanga pūnoa, ina ko x^{2}+bx+c he pūrua tika pūrua tika pū, ka taea taua mea te tauwehea i ngā wā katoa hei \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{5}{3}\right)^{2}=\frac{49}{9}$$

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

$$\sqrt{\left(x+\frac{5}{3}\right)^{2}}=\sqrt{\frac{49}{9}}$$

Whakarūnātia.

$$x+\frac{5}{3}=\frac{7}{3} x+\frac{5}{3}=-\frac{7}{3}$$

Me tango \frac{5}{3} mai i ngā taha e rua o te whārite.

$$x=\frac{2}{3} x=-4$$