Whakaoti i te x(-11x/5)+5(-11x/5)=22

$$x(- \frac{ 11x }{ 5 } )+5(- \frac{ 11x }{ 5 } )=22$$

Whakaoti mō x (complex solution)

$$x=\frac{-5+\sqrt{15}i}{2}\approx -2.5+1.936491673i$$

$$x=\frac{-\sqrt{15}i-5}{2}\approx -2.5-1.936491673i$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakareatia ngā taha e rua o te whārite ki te 5.

$$5x\left(-\frac{11x}{5}\right)+25\left(-\frac{11x}{5}\right)=110$$

Tuhia te 5\left(-\frac{11x}{5}\right) hei hautanga kotahi.

$$\frac{-5\times 11x}{5}x+25\left(-\frac{11x}{5}\right)=110$$

Me whakakore te 5 me te 5.

$$-11xx+25\left(-\frac{11x}{5}\right)=110$$

Whakakorea atu te tauwehe pūnoa nui rawa 5 i roto i te 25 me te 5.

$$-11xx-5\times 11x=110$$

Whakareatia te -1 ki te 11, ka -11. Whakareatia te -5 ki te 11, ka -55.

$$-11xx-55x=110$$

Whakareatia te x ki te x, ka x^{2}.

$$-11x^{2}-55x=110$$

Tangohia te 110 mai i ngā taha e rua.

$$-11x^{2}-55x-110=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi -11 mō a, -55 mō b, me -110 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-55\right)±\sqrt{\left(-55\right)^{2}-4\left(-11\right)\left(-110\right)}}{2\left(-11\right)}$$

Pūrua -55.

$$x=\frac{-\left(-55\right)±\sqrt{3025-4\left(-11\right)\left(-110\right)}}{2\left(-11\right)}$$

Whakareatia -4 ki te -11.

$$x=\frac{-\left(-55\right)±\sqrt{3025+44\left(-110\right)}}{2\left(-11\right)}$$

Whakareatia 44 ki te -110.

$$x=\frac{-\left(-55\right)±\sqrt{3025-4840}}{2\left(-11\right)}$$

Tāpiri 3025 ki te -4840.

$$x=\frac{-\left(-55\right)±\sqrt{-1815}}{2\left(-11\right)}$$

Tuhia te pūtakerua o te -1815.

$$x=\frac{-\left(-55\right)±11\sqrt{15}i}{2\left(-11\right)}$$

Ko te tauaro o -55 ko 55.

$$x=\frac{55±11\sqrt{15}i}{2\left(-11\right)}$$

Whakareatia 2 ki te -11.

$$x=\frac{55±11\sqrt{15}i}{-22}$$

Nā, me whakaoti te whārite x=\frac{55±11\sqrt{15}i}{-22} ina he tāpiri te ±. Tāpiri 55 ki te 11i\sqrt{15}.

$$x=\frac{55+11\sqrt{15}i}{-22}$$

Whakawehe 55+11i\sqrt{15} ki te -22.

$$x=\frac{-\sqrt{15}i-5}{2}$$

Nā, me whakaoti te whārite x=\frac{55±11\sqrt{15}i}{-22} ina he tango te ±. Tango 11i\sqrt{15} mai i 55.

$$x=\frac{-11\sqrt{15}i+55}{-22}$$

Whakawehe 55-11i\sqrt{15} ki te -22.

$$x=\frac{-5+\sqrt{15}i}{2}$$

Kua oti te whārite te whakatau.

$$x=\frac{-\sqrt{15}i-5}{2} x=\frac{-5+\sqrt{15}i}{2}$$

Ngā Upane mō te Whakaoti i te Pūrua

Whakareatia ngā taha e rua o te whārite ki te 5.

$$5x\left(-\frac{11x}{5}\right)+25\left(-\frac{11x}{5}\right)=110$$

Tuhia te 5\left(-\frac{11x}{5}\right) hei hautanga kotahi.

$$\frac{-5\times 11x}{5}x+25\left(-\frac{11x}{5}\right)=110$$

Me whakakore te 5 me te 5.

$$-11xx+25\left(-\frac{11x}{5}\right)=110$$

Whakakorea atu te tauwehe pūnoa nui rawa 5 i roto i te 25 me te 5.

$$-11xx-5\times 11x=110$$

Whakareatia te -1 ki te 11, ka -11. Whakareatia te -5 ki te 11, ka -55.

$$-11xx-55x=110$$

Whakareatia te x ki te x, ka x^{2}.

$$-11x^{2}-55x=110$$

Whakawehea ngā taha e rua ki te -11.

$$\frac{-11x^{2}-55x}{-11}=\frac{110}{-11}$$

Mā te whakawehe ki te -11 ka wetekia te whakareanga ki te -11.

$$x^{2}+\left(-\frac{55}{-11}\right)x=\frac{110}{-11}$$

Whakawehe -55 ki te -11.

$$x^{2}+5x=\frac{110}{-11}$$

Whakawehe 110 ki te -11.

$$x^{2}+5x=-10$$

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.

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

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

$$x^{2}+5x+\frac{25}{4}=-10+\frac{25}{4}$$

Tāpiri -10 ki te \frac{25}{4}.

$$x^{2}+5x+\frac{25}{4}=-\frac{15}{4}$$

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}.

$$\left(x+\frac{5}{2}\right)^{2}=-\frac{15}{4}$$

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

$$\sqrt{\left(x+\frac{5}{2}\right)^{2}}=\sqrt{-\frac{15}{4}}$$

Whakarūnātia.

$$x+\frac{5}{2}=\frac{\sqrt{15}i}{2} x+\frac{5}{2}=-\frac{\sqrt{15}i}{2}$$

Me tango \frac{5}{2} mai i ngā taha e rua o te whārite.

$$x=\frac{-5+\sqrt{15}i}{2} x=\frac{-\sqrt{15}i-5}{2}$$