Whakaoti i te (k+1/4)^2-1/16-1/5=0

$$( k + \frac { 1 } { 4 } ) ^ { 2 } - \frac { 1 } { 16 } - \frac { 1 } { 5 } = 0$$

Whakaoti mō k

$$k=\frac{\sqrt{105}}{20}-\frac{1}{4}\approx 0.262347538$$

$$k=-\frac{\sqrt{105}}{20}-\frac{1}{4}\approx -0.762347538$$

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(k+\frac{1}{4}\right)^{2}.

$$k^{2}+\frac{1}{2}k+\frac{1}{16}-\frac{1}{16}-\frac{1}{5}=0$$

Tangohia te \frac{1}{16} i te \frac{1}{16}, ka 0.

$$k^{2}+\frac{1}{2}k-\frac{1}{5}=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, \frac{1}{2} mō b, me -\frac{1}{5} mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$k=\frac{-\frac{1}{2}±\sqrt{\left(\frac{1}{2}\right)^{2}-4\left(-\frac{1}{5}\right)}}{2}$$

Pūruatia \frac{1}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$k=\frac{-\frac{1}{2}±\sqrt{\frac{1}{4}-4\left(-\frac{1}{5}\right)}}{2}$$

Whakareatia -4 ki te -\frac{1}{5}.

$$k=\frac{-\frac{1}{2}±\sqrt{\frac{1}{4}+\frac{4}{5}}}{2}$$

Tāpiri \frac{1}{4} ki te \frac{4}{5} 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.

$$k=\frac{-\frac{1}{2}±\sqrt{\frac{21}{20}}}{2}$$

Tuhia te pūtakerua o te \frac{21}{20}.

$$k=\frac{-\frac{1}{2}±\frac{\sqrt{105}}{10}}{2}$$

Nā, me whakaoti te whārite k=\frac{-\frac{1}{2}±\frac{\sqrt{105}}{10}}{2} ina he tāpiri te ±. Tāpiri -\frac{1}{2} ki te \frac{\sqrt{105}}{10}.

$$k=\frac{\frac{\sqrt{105}}{10}-\frac{1}{2}}{2}$$

Whakawehe -\frac{1}{2}+\frac{\sqrt{105}}{10} ki te 2.

$$k=\frac{\sqrt{105}}{20}-\frac{1}{4}$$

Nā, me whakaoti te whārite k=\frac{-\frac{1}{2}±\frac{\sqrt{105}}{10}}{2} ina he tango te ±. Tango \frac{\sqrt{105}}{10} mai i -\frac{1}{2}.

$$k=\frac{-\frac{\sqrt{105}}{10}-\frac{1}{2}}{2}$$

Whakawehe -\frac{1}{2}-\frac{\sqrt{105}}{10} ki te 2.

$$k=-\frac{\sqrt{105}}{20}-\frac{1}{4}$$

Kua oti te whārite te whakatau.

$$k=\frac{\sqrt{105}}{20}-\frac{1}{4} k=-\frac{\sqrt{105}}{20}-\frac{1}{4}$$

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(k+\frac{1}{4}\right)^{2}.

$$k^{2}+\frac{1}{2}k+\frac{1}{16}-\frac{1}{16}-\frac{1}{5}=0$$

Tangohia te \frac{1}{16} i te \frac{1}{16}, ka 0.

$$k^{2}+\frac{1}{2}k-\frac{1}{5}=0$$

Me tāpiri te \frac{1}{5} ki ngā taha e rua. Ko te tau i tāpiria he kore ka hua koia tonu.

$$k^{2}+\frac{1}{2}k=\frac{1}{5}$$

Whakawehea te \frac{1}{2}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{1}{4}. Nā, tāpiria te pūrua o te \frac{1}{4} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$k^{2}+\frac{1}{2}k+\left(\frac{1}{4}\right)^{2}=\frac{1}{5}+\left(\frac{1}{4}\right)^{2}$$

Pūruatia \frac{1}{4} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$k^{2}+\frac{1}{2}k+\frac{1}{16}=\frac{1}{5}+\frac{1}{16}$$

Tāpiri \frac{1}{5} ki te \frac{1}{16} 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.

$$k^{2}+\frac{1}{2}k+\frac{1}{16}=\frac{21}{80}$$

Tauwehea k^{2}+\frac{1}{2}k+\frac{1}{16}. 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}.

$$\left(k+\frac{1}{4}\right)^{2}=\frac{21}{80}$$

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

$$\sqrt{\left(k+\frac{1}{4}\right)^{2}}=\sqrt{\frac{21}{80}}$$

Whakarūnātia.

$$k+\frac{1}{4}=\frac{\sqrt{105}}{20} k+\frac{1}{4}=-\frac{\sqrt{105}}{20}$$

Me tango \frac{1}{4} mai i ngā taha e rua o te whārite.

$$k=\frac{\sqrt{105}}{20}-\frac{1}{4} k=-\frac{\sqrt{105}}{20}-\frac{1}{4}$$