Whakaoti i te 3^2=2.5^2+x^2

$${ 3 }^{ 2 } = { 2.5 }^{ 2 } + { x }^{ 2 }$$

Whakaoti mō x

$$x = \frac{\sqrt{11}}{2} \approx 1.658312395$$

$$x = -\frac{\sqrt{11}}{2} \approx -1.658312395$$

Ngā Upane mā te Kimi Pūtakerua

Tātaihia te 3 mā te pū o 2, kia riro ko 9.

$$9=2.5^{2}+x^{2}$$

Tātaihia te 2.5 mā te pū o 2, kia riro ko 6.25.

$$9=6.25+x^{2}$$

Whakawhitihia ngā taha kia puta ki te taha mauī ngā kīanga tau taurangi katoa.

$$6.25+x^{2}=9$$

Tangohia te 6.25 mai i ngā taha e rua.

$$x^{2}=9-6.25$$

Tangohia te 6.25 i te 9, ka 2.75.

$$x^{2}=2.75$$

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

$$x=\frac{\sqrt{11}}{2} x=-\frac{\sqrt{11}}{2}$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tātaihia te 3 mā te pū o 2, kia riro ko 9.

$$9=2.5^{2}+x^{2}$$

Tātaihia te 2.5 mā te pū o 2, kia riro ko 6.25.

$$9=6.25+x^{2}$$

Whakawhitihia ngā taha kia puta ki te taha mauī ngā kīanga tau taurangi katoa.

$$6.25+x^{2}=9$$

Tangohia te 9 mai i ngā taha e rua.

$$6.25+x^{2}-9=0$$

Tangohia te 9 i te 6.25, ka -2.75.

$$-2.75+x^{2}=0$$

Ko ngā tikanga tātai pūrua pēnei i tēnei nā, me te kīanga tau x^{2} engari kāore he kīanga tau x, ka taea tonu te whakaoti mā te whakamahi i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}, ina tuhia ki te tānga ngahuru: ax^{2}+bx+c=0.

$$x^{2}-2.75=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, 0 mō b, me -2.75 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-2.75\right)}}{2}$$

Pūrua 0.

$$x=\frac{0±\sqrt{-4\left(-2.75\right)}}{2}$$

Whakareatia -4 ki te -2.75.

$$x=\frac{0±\sqrt{11}}{2}$$

Nā, me whakaoti te whārite x=\frac{0±\sqrt{11}}{2} ina he tāpiri te ±.

$$x=\frac{\sqrt{11}}{2}$$

Nā, me whakaoti te whārite x=\frac{0±\sqrt{11}}{2} ina he tango te ±.

$$x=-\frac{\sqrt{11}}{2}$$

Kua oti te whārite te whakatau.

$$x=\frac{\sqrt{11}}{2} x=-\frac{\sqrt{11}}{2}$$