Whakaoti i te left(5x+1right)^2-3(5x+1)-4=0

$${ \left(5x+1 \right) }^{ 2 } -3(5x+1)-4=0$$

Whakaoti mō x

$$x=-\frac{2}{5}=-0.4$$

$$x=\frac{3}{5}=0.6$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

Whakamahia te ture huarua \left(a+b\right)^{2}=a^{2}+2ab+b^{2} hei whakaroha \left(5x+1\right)^{2}.

$$25x^{2}+10x+1-3\left(5x+1\right)-4=0$$

Whakamahia te āhuatanga tohatoha hei whakarea te -3 ki te 5x+1.

$$25x^{2}+10x+1-15x-3-4=0$$

Pahekotia te 10x me -15x, ka -5x.

$$25x^{2}-5x+1-3-4=0$$

Tangohia te 3 i te 1, ka -2.

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

Tangohia te 4 i te -2, ka -6.

$$25x^{2}-5x-6=0$$

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei 25x^{2}+ax+bx-6. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-5 ab=25\left(-6\right)=-150$$

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 -150.

$$1,-150 2,-75 3,-50 5,-30 6,-25 10,-15$$

Tātaihia te tapeke mō ia takirua.

$$1-150=-149 2-75=-73 3-50=-47 5-30=-25 6-25=-19 10-15=-5$$

Ko te otinga te takirua ka hoatu i te tapeke -5.

$$a=-15 b=10$$

Tuhia anō te 25x^{2}-5x-6 hei \left(25x^{2}-15x\right)+\left(10x-6\right).

$$\left(25x^{2}-15x\right)+\left(10x-6\right)$$

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

$$5x\left(5x-3\right)+2\left(5x-3\right)$$

Whakatauwehea atu te kīanga pātahi 5x-3 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(5x-3\right)\left(5x+2\right)$$

Hei kimi otinga whārite, me whakaoti te 5x-3=0 me te 5x+2=0.

$$x=\frac{3}{5} x=-\frac{2}{5}$$

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(5x+1\right)^{2}.

$$25x^{2}+10x+1-3\left(5x+1\right)-4=0$$

Whakamahia te āhuatanga tohatoha hei whakarea te -3 ki te 5x+1.

$$25x^{2}+10x+1-15x-3-4=0$$

Pahekotia te 10x me -15x, ka -5x.

$$25x^{2}-5x+1-3-4=0$$

Tangohia te 3 i te 1, ka -2.

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

Tangohia te 4 i te -2, ka -6.

$$25x^{2}-5x-6=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 25 mō a, -5 mō b, me -6 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 25\left(-6\right)}}{2\times 25}$$

Pūrua -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 25\left(-6\right)}}{2\times 25}$$

Whakareatia -4 ki te 25.

$$x=\frac{-\left(-5\right)±\sqrt{25-100\left(-6\right)}}{2\times 25}$$

Whakareatia -100 ki te -6.

$$x=\frac{-\left(-5\right)±\sqrt{25+600}}{2\times 25}$$

Tāpiri 25 ki te 600.

$$x=\frac{-\left(-5\right)±\sqrt{625}}{2\times 25}$$

Tuhia te pūtakerua o te 625.

$$x=\frac{-\left(-5\right)±25}{2\times 25}$$

Ko te tauaro o -5 ko 5.

$$x=\frac{5±25}{2\times 25}$$

Whakareatia 2 ki te 25.

$$x=\frac{5±25}{50}$$

Nā, me whakaoti te whārite x=\frac{5±25}{50} ina he tāpiri te ±. Tāpiri 5 ki te 25.

$$x=\frac{30}{50}$$

Whakahekea te hautanga \frac{30}{50} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 10.

$$x=\frac{3}{5}$$

Nā, me whakaoti te whārite x=\frac{5±25}{50} ina he tango te ±. Tango 25 mai i 5.

$$x=-\frac{20}{50}$$

Whakahekea te hautanga \frac{-20}{50} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 10.

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

Kua oti te whārite te whakatau.

$$x=\frac{3}{5} x=-\frac{2}{5}$$

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(5x+1\right)^{2}.

$$25x^{2}+10x+1-3\left(5x+1\right)-4=0$$

Whakamahia te āhuatanga tohatoha hei whakarea te -3 ki te 5x+1.

$$25x^{2}+10x+1-15x-3-4=0$$

Pahekotia te 10x me -15x, ka -5x.

$$25x^{2}-5x+1-3-4=0$$

Tangohia te 3 i te 1, ka -2.

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

Tangohia te 4 i te -2, ka -6.

$$25x^{2}-5x-6=0$$

Me tāpiri te 6 ki ngā taha e rua. Ko te tau i tāpiria he kore ka hua koia tonu.

$$25x^{2}-5x=6$$

Whakawehea ngā taha e rua ki te 25.

$$\frac{25x^{2}-5x}{25}=\frac{6}{25}$$

Mā te whakawehe ki te 25 ka wetekia te whakareanga ki te 25.

$$x^{2}+\left(-\frac{5}{25}\right)x=\frac{6}{25}$$

Whakahekea te hautanga \frac{-5}{25} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 5.

$$x^{2}-\frac{1}{5}x=\frac{6}{25}$$

Whakawehea te -\frac{1}{5}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{1}{10}. Nā, tāpiria te pūrua o te -\frac{1}{10} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}-\frac{1}{5}x+\left(-\frac{1}{10}\right)^{2}=\frac{6}{25}+\left(-\frac{1}{10}\right)^{2}$$

Pūruatia -\frac{1}{10} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-\frac{1}{5}x+\frac{1}{100}=\frac{6}{25}+\frac{1}{100}$$

Tāpiri \frac{6}{25} ki te \frac{1}{100} 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.

$$x^{2}-\frac{1}{5}x+\frac{1}{100}=\frac{1}{4}$$

Tauwehea x^{2}-\frac{1}{5}x+\frac{1}{100}. 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}.

$$\left(x-\frac{1}{10}\right)^{2}=\frac{1}{4}$$

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

$$\sqrt{\left(x-\frac{1}{10}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Whakarūnātia.

$$x-\frac{1}{10}=\frac{1}{2} x-\frac{1}{10}=-\frac{1}{2}$$

Me tāpiri \frac{1}{10} ki ngā taha e rua o te whārite.

$$x=\frac{3}{5} x=-\frac{2}{5}$$