Whakaoti i te x^2+(5-x)^2=(5-2x)^2

$$x ^ { 2 } + ( 5 - x ) ^ { 2 } = ( 5 - 2 x ) ^ { 2 }$$

Whakaoti mō x

$$x=5$$

$$x=0$$

Ngā Upane Whakamahi Tauwehe

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

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

Pahekotia te x^{2} me x^{2}, ka 2x^{2}.

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

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

$$2x^{2}+25-10x=25-20x+4x^{2}$$

Tangohia te 25 mai i ngā taha e rua.

$$2x^{2}+25-10x-25=-20x+4x^{2}$$

Tangohia te 25 i te 25, ka 0.

$$2x^{2}-10x=-20x+4x^{2}$$

Me tāpiri te 20x ki ngā taha e rua.

$$2x^{2}-10x+20x=4x^{2}$$

Pahekotia te -10x me 20x, ka 10x.

$$2x^{2}+10x=4x^{2}$$

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

$$2x^{2}+10x-4x^{2}=0$$

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

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

Tauwehea te x.

$$x\left(-2x+10\right)=0$$

Hei kimi otinga whārite, me whakaoti te x=0 me te -2x+10=0.

$$x=0 x=5$$

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

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

Pahekotia te x^{2} me x^{2}, ka 2x^{2}.

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

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

$$2x^{2}+25-10x=25-20x+4x^{2}$$

Tangohia te 25 mai i ngā taha e rua.

$$2x^{2}+25-10x-25=-20x+4x^{2}$$

Tangohia te 25 i te 25, ka 0.

$$2x^{2}-10x=-20x+4x^{2}$$

Me tāpiri te 20x ki ngā taha e rua.

$$2x^{2}-10x+20x=4x^{2}$$

Pahekotia te -10x me 20x, ka 10x.

$$2x^{2}+10x=4x^{2}$$

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

$$2x^{2}+10x-4x^{2}=0$$

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

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

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

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

Tuhia te pūtakerua o te 10^{2}.

$$x=\frac{-10±10}{2\left(-2\right)}$$

Whakareatia 2 ki te -2.

$$x=\frac{-10±10}{-4}$$

Nā, me whakaoti te whārite x=\frac{-10±10}{-4} ina he tāpiri te ±. Tāpiri -10 ki te 10.

$$x=\frac{0}{-4}$$

Whakawehe 0 ki te -4.

$$x=0$$

Nā, me whakaoti te whārite x=\frac{-10±10}{-4} ina he tango te ±. Tango 10 mai i -10.

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

Whakawehe -20 ki te -4.

$$x=5$$

Kua oti te whārite te whakatau.

$$x=0 x=5$$

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

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

Pahekotia te x^{2} me x^{2}, ka 2x^{2}.

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

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

$$2x^{2}+25-10x=25-20x+4x^{2}$$

Me tāpiri te 20x ki ngā taha e rua.

$$2x^{2}+25-10x+20x=25+4x^{2}$$

Pahekotia te -10x me 20x, ka 10x.

$$2x^{2}+25+10x=25+4x^{2}$$

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

$$2x^{2}+25+10x-4x^{2}=25$$

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

$$-2x^{2}+25+10x=25$$

Tangohia te 25 mai i ngā taha e rua.

$$-2x^{2}+10x=25-25$$

Tangohia te 25 i te 25, ka 0.

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

Whakawehea ngā taha e rua ki te -2.

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

Mā te whakawehe ki te -2 ka wetekia te whakareanga ki te -2.

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

Whakawehe 10 ki te -2.

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

Whakawehe 0 ki te -2.

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

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.

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

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

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

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}.

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

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

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

Whakarūnātia.

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

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

$$x=5 x=0$$