Whakaoti i te 1(2x-5)^2-09(x+4)^2=0

$$1 ( 2 x - 5 ) ^ { 2 } - 09 ( x + 4 ) ^ { 2 } = 0$$

Whakaoti mō x

$$x = \frac{5}{2} = 2\frac{1}{2} = 2.5$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(2x-5\right)^{2}.

$$1\left(4x^{2}-20x+25\right)-0\times 9\left(x+4\right)^{2}=0$$

Whakamahia te āhuatanga tohatoha hei whakarea te 1 ki te 4x^{2}-20x+25.

$$4x^{2}-20x+25-0\times 9\left(x+4\right)^{2}=0$$

Whakareatia te 0 ki te 9, ka 0.

$$4x^{2}-20x+25-0\left(x+4\right)^{2}=0$$

Whakamahia te ture huarua \left(a+b\right)^{2}=a^{2}+2ab+b^{2} hei whakaroha \left(x+4\right)^{2}.

$$4x^{2}-20x+25-0\left(x^{2}+8x+16\right)=0$$

Ko te tau i whakarea ki te kore ka hua ko te kore.

$$4x^{2}-20x+25-0=0$$

Whakaraupapatia anō ngā kīanga tau.

$$4x^{2}-20x+25=0$$

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei 4x^{2}+ax+bx+25. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-20 ab=4\times 25=100$$

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 100.

$$-1,-100 -2,-50 -4,-25 -5,-20 -10,-10$$

Tātaihia te tapeke mō ia takirua.

$$-1-100=-101 -2-50=-52 -4-25=-29 -5-20=-25 -10-10=-20$$

Ko te otinga te takirua ka hoatu i te tapeke -20.

$$a=-10 b=-10$$

Tuhia anō te 4x^{2}-20x+25 hei \left(4x^{2}-10x\right)+\left(-10x+25\right).

$$\left(4x^{2}-10x\right)+\left(-10x+25\right)$$

Tauwehea te 2x i te tuatahi me te -5 i te rōpū tuarua.

$$2x\left(2x-5\right)-5\left(2x-5\right)$$

Whakatauwehea atu te kīanga pātahi 2x-5 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(2x-5\right)\left(2x-5\right)$$

Tuhia anōtia hei pūrua huarua.

$$\left(2x-5\right)^{2}$$

Hei kimi i te otinga whārite, whakaotia te 2x-5=0.

$$x=\frac{5}{2}$$

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(2x-5\right)^{2}.

$$1\left(4x^{2}-20x+25\right)-0\times 9\left(x+4\right)^{2}=0$$

Whakamahia te āhuatanga tohatoha hei whakarea te 1 ki te 4x^{2}-20x+25.

$$4x^{2}-20x+25-0\times 9\left(x+4\right)^{2}=0$$

Whakareatia te 0 ki te 9, ka 0.

$$4x^{2}-20x+25-0\left(x+4\right)^{2}=0$$

Whakamahia te ture huarua \left(a+b\right)^{2}=a^{2}+2ab+b^{2} hei whakaroha \left(x+4\right)^{2}.

$$4x^{2}-20x+25-0\left(x^{2}+8x+16\right)=0$$

Ko te tau i whakarea ki te kore ka hua ko te kore.

$$4x^{2}-20x+25-0=0$$

Whakaraupapatia anō ngā kīanga tau.

$$4x^{2}-20x+25=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 4 mō a, -20 mō b, me 25 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times 4\times 25}}{2\times 4}$$

Pūrua -20.

$$x=\frac{-\left(-20\right)±\sqrt{400-4\times 4\times 25}}{2\times 4}$$

Whakareatia -4 ki te 4.

$$x=\frac{-\left(-20\right)±\sqrt{400-16\times 25}}{2\times 4}$$

Whakareatia -16 ki te 25.

$$x=\frac{-\left(-20\right)±\sqrt{400-400}}{2\times 4}$$

Tāpiri 400 ki te -400.

$$x=\frac{-\left(-20\right)±\sqrt{0}}{2\times 4}$$

Tuhia te pūtakerua o te 0.

$$x=-\frac{-20}{2\times 4}$$

Ko te tauaro o -20 ko 20.

$$x=\frac{20}{2\times 4}$$

Whakareatia 2 ki te 4.

$$x=\frac{20}{8}$$

Whakahekea te hautanga \frac{20}{8} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 4.

$$x=\frac{5}{2}$$

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(2x-5\right)^{2}.

$$1\left(4x^{2}-20x+25\right)-0\times 9\left(x+4\right)^{2}=0$$

Whakamahia te āhuatanga tohatoha hei whakarea te 1 ki te 4x^{2}-20x+25.

$$4x^{2}-20x+25-0\times 9\left(x+4\right)^{2}=0$$

Whakareatia te 0 ki te 9, ka 0.

$$4x^{2}-20x+25-0\left(x+4\right)^{2}=0$$

Whakamahia te ture huarua \left(a+b\right)^{2}=a^{2}+2ab+b^{2} hei whakaroha \left(x+4\right)^{2}.

$$4x^{2}-20x+25-0\left(x^{2}+8x+16\right)=0$$

Ko te tau i whakarea ki te kore ka hua ko te kore.

$$4x^{2}-20x+25-0=0$$

Me tāpiri te 0 ki ngā taha e rua.

$$4x^{2}-20x+25=0+0$$

Tāpirihia te 0 ki te 0, ka 0.

$$4x^{2}-20x+25=0$$

Tangohia te 25 mai i ngā taha e rua. Ko te tau i tango i te kore ka hua ko tōna korenga.

$$4x^{2}-20x=-25$$

Whakawehea ngā taha e rua ki te 4.

$$\frac{4x^{2}-20x}{4}=-\frac{25}{4}$$

Mā te whakawehe ki te 4 ka wetekia te whakareanga ki te 4.

$$x^{2}+\left(-\frac{20}{4}\right)x=-\frac{25}{4}$$

Whakawehe -20 ki te 4.

$$x^{2}-5x=-\frac{25}{4}$$

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

$$x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=-\frac{25}{4}+\left(-\frac{5}{2}\right)^{2}$$

Pūruatia -\frac{5}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-5x+\frac{25}{4}=\frac{-25+25}{4}$$

Tāpiri -\frac{25}{4} ki te \frac{25}{4} 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.

$$x^{2}-5x+\frac{25}{4}=0$$

Tauwehea x^{2}-5x+\frac{25}{4}. 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}.

$$\left(x-\frac{5}{2}\right)^{2}=0$$

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

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{0}$$

Whakarūnātia.

$$x-\frac{5}{2}=0 x-\frac{5}{2}=0$$

Me tāpiri \frac{5}{2} ki ngā taha e rua o te whārite.

$$x=\frac{5}{2} x=\frac{5}{2}$$

Kua oti te whārite te whakatau. He ōrite ngā whakatau.

$$x=\frac{5}{2}$$