Whakaoti i te 2=-1/4x^2+5/2x

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

Whakaoti mō x

$$x=\sqrt{17}+5\approx 9.123105626$$

$$x=5-\sqrt{17}\approx 0.876894374$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

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

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

Tangohia te 2 mai i ngā taha e rua.

$$-\frac{1}{4}x^{2}+\frac{5}{2}x-2=0$$

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

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

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

$$x=\frac{-\frac{5}{2}±\sqrt{\frac{25}{4}-4\left(-\frac{1}{4}\right)\left(-2\right)}}{2\left(-\frac{1}{4}\right)}$$

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

$$x=\frac{-\frac{5}{2}±\sqrt{\frac{25}{4}-2}}{2\left(-\frac{1}{4}\right)}$$

Tāpiri \frac{25}{4} ki te -2.

$$x=\frac{-\frac{5}{2}±\sqrt{\frac{17}{4}}}{2\left(-\frac{1}{4}\right)}$$

Tuhia te pūtakerua o te \frac{17}{4}.

$$x=\frac{-\frac{5}{2}±\frac{\sqrt{17}}{2}}{2\left(-\frac{1}{4}\right)}$$

Whakareatia 2 ki te -\frac{1}{4}.

$$x=\frac{-\frac{5}{2}±\frac{\sqrt{17}}{2}}{-\frac{1}{2}}$$

Nā, me whakaoti te whārite x=\frac{-\frac{5}{2}±\frac{\sqrt{17}}{2}}{-\frac{1}{2}} ina he tāpiri te ±. Tāpiri -\frac{5}{2} ki te \frac{\sqrt{17}}{2}.

$$x=\frac{\sqrt{17}-5}{-\frac{1}{2}\times 2}$$

Whakawehe \frac{-5+\sqrt{17}}{2} ki te -\frac{1}{2} mā te whakarea \frac{-5+\sqrt{17}}{2} ki te tau huripoki o -\frac{1}{2}.

$$x=5-\sqrt{17}$$

Nā, me whakaoti te whārite x=\frac{-\frac{5}{2}±\frac{\sqrt{17}}{2}}{-\frac{1}{2}} ina he tango te ±. Tango \frac{\sqrt{17}}{2} mai i -\frac{5}{2}.

$$x=\frac{-\sqrt{17}-5}{-\frac{1}{2}\times 2}$$

Whakawehe \frac{-5-\sqrt{17}}{2} ki te -\frac{1}{2} mā te whakarea \frac{-5-\sqrt{17}}{2} ki te tau huripoki o -\frac{1}{2}.

$$x=\sqrt{17}+5$$

Kua oti te whārite te whakatau.

$$x=5-\sqrt{17} x=\sqrt{17}+5$$

Ngā Upane mō te Whakaoti i te Pūrua

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

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

Me whakarea ngā taha e rua ki te -4.

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

Mā te whakawehe ki te -\frac{1}{4} ka wetekia te whakareanga ki te -\frac{1}{4}.

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

Whakawehe \frac{5}{2} ki te -\frac{1}{4} mā te whakarea \frac{5}{2} ki te tau huripoki o -\frac{1}{4}.

$$x^{2}-10x=\frac{2}{-\frac{1}{4}}$$

Whakawehe 2 ki te -\frac{1}{4} mā te whakarea 2 ki te tau huripoki o -\frac{1}{4}.

$$x^{2}-10x=-8$$

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

$$x^{2}-10x+\left(-5\right)^{2}=-8+\left(-5\right)^{2}$$

Pūrua -5.

$$x^{2}-10x+25=-8+25$$

Tāpiri -8 ki te 25.

$$x^{2}-10x+25=17$$

Tauwehea x^{2}-10x+25. 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}.

$$\left(x-5\right)^{2}=17$$

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

$$\sqrt{\left(x-5\right)^{2}}=\sqrt{17}$$

Whakarūnātia.

$$x-5=\sqrt{17} x-5=-\sqrt{17}$$

Me tāpiri 5 ki ngā taha e rua o te whārite.

$$x=\sqrt{17}+5 x=5-\sqrt{17}$$