Whakaoti i te {left(1/4xright)}^2+{left(80/4-1/4xright)}^2=200

$${ \left( \frac{ 1 }{ 4 } x \right) }^{ 2 } + { \left( \frac{ 80 }{ 4 } - \frac{ 1 }{ 4 } x \right) }^{ 2 } =200$$

Whakaoti mō x

$$x=40$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakarohaina te \left(\frac{1}{4}x\right)^{2}.

$$\left(\frac{1}{4}\right)^{2}x^{2}+\left(\frac{80}{4}-\frac{1}{4}x\right)^{2}=200$$

Tātaihia te \frac{1}{4} mā te pū o 2, kia riro ko \frac{1}{16}.

$$\frac{1}{16}x^{2}+\left(\frac{80}{4}-\frac{1}{4}x\right)^{2}=200$$

Whakawehea te 80 ki te 4, kia riro ko 20.

$$\frac{1}{16}x^{2}+\left(20-\frac{1}{4}x\right)^{2}=200$$

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(20-\frac{1}{4}x\right)^{2}.

$$\frac{1}{16}x^{2}+400-10x+\frac{1}{16}x^{2}=200$$

Pahekotia te \frac{1}{16}x^{2} me \frac{1}{16}x^{2}, ka \frac{1}{8}x^{2}.

$$\frac{1}{8}x^{2}+400-10x=200$$

Tangohia te 200 mai i ngā taha e rua.

$$\frac{1}{8}x^{2}+400-10x-200=0$$

Tangohia te 200 i te 400, ka 200.

$$\frac{1}{8}x^{2}+200-10x=0$$

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.

$$\frac{1}{8}x^{2}-10x+200=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi \frac{1}{8} mō a, -10 mō b, me 200 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times \frac{1}{8}\times 200}}{2\times \frac{1}{8}}$$

Pūrua -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\times \frac{1}{8}\times 200}}{2\times \frac{1}{8}}$$

Whakareatia -4 ki te \frac{1}{8}.

$$x=\frac{-\left(-10\right)±\sqrt{100-\frac{1}{2}\times 200}}{2\times \frac{1}{8}}$$

Whakareatia -\frac{1}{2} ki te 200.

$$x=\frac{-\left(-10\right)±\sqrt{100-100}}{2\times \frac{1}{8}}$$

Tāpiri 100 ki te -100.

$$x=\frac{-\left(-10\right)±\sqrt{0}}{2\times \frac{1}{8}}$$

Tuhia te pūtakerua o te 0.

$$x=-\frac{-10}{2\times \frac{1}{8}}$$

Ko te tauaro o -10 ko 10.

$$x=\frac{10}{2\times \frac{1}{8}}$$

Whakareatia 2 ki te \frac{1}{8}.

$$x=\frac{10}{\frac{1}{4}}$$

Whakawehe 10 ki te \frac{1}{4} mā te whakarea 10 ki te tau huripoki o \frac{1}{4}.

$$x=40$$

Ngā Upane mō te Whakaoti i te Pūrua

Whakarohaina te \left(\frac{1}{4}x\right)^{2}.

$$\left(\frac{1}{4}\right)^{2}x^{2}+\left(\frac{80}{4}-\frac{1}{4}x\right)^{2}=200$$

Tātaihia te \frac{1}{4} mā te pū o 2, kia riro ko \frac{1}{16}.

$$\frac{1}{16}x^{2}+\left(\frac{80}{4}-\frac{1}{4}x\right)^{2}=200$$

Whakawehea te 80 ki te 4, kia riro ko 20.

$$\frac{1}{16}x^{2}+\left(20-\frac{1}{4}x\right)^{2}=200$$

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(20-\frac{1}{4}x\right)^{2}.

$$\frac{1}{16}x^{2}+400-10x+\frac{1}{16}x^{2}=200$$

Pahekotia te \frac{1}{16}x^{2} me \frac{1}{16}x^{2}, ka \frac{1}{8}x^{2}.

$$\frac{1}{8}x^{2}+400-10x=200$$

Tangohia te 400 mai i ngā taha e rua.

$$\frac{1}{8}x^{2}-10x=200-400$$

Tangohia te 400 i te 200, ka -200.

$$\frac{1}{8}x^{2}-10x=-200$$

Me whakarea ngā taha e rua ki te 8.

$$\frac{\frac{1}{8}x^{2}-10x}{\frac{1}{8}}=-\frac{200}{\frac{1}{8}}$$

Mā te whakawehe ki te \frac{1}{8} ka wetekia te whakareanga ki te \frac{1}{8}.

$$x^{2}+\left(-\frac{10}{\frac{1}{8}}\right)x=-\frac{200}{\frac{1}{8}}$$

Whakawehe -10 ki te \frac{1}{8} mā te whakarea -10 ki te tau huripoki o \frac{1}{8}.

$$x^{2}-80x=-\frac{200}{\frac{1}{8}}$$

Whakawehe -200 ki te \frac{1}{8} mā te whakarea -200 ki te tau huripoki o \frac{1}{8}.

$$x^{2}-80x=-1600$$

Whakawehea te -80, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -40. Nā, tāpiria te pūrua o te -40 ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}-80x+\left(-40\right)^{2}=-1600+\left(-40\right)^{2}$$

Pūrua -40.

$$x^{2}-80x+1600=-1600+1600$$

Tāpiri -1600 ki te 1600.

$$x^{2}-80x+1600=0$$

Tauwehea x^{2}-80x+1600. 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}.

$$\left(x-40\right)^{2}=0$$

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

$$\sqrt{\left(x-40\right)^{2}}=\sqrt{0}$$

Whakarūnātia.

$$x-40=0 x-40=0$$

Me tāpiri 40 ki ngā taha e rua o te whārite.

$$x=40 x=40$$

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

$$x=40$$