Whakaoti i te (x+5)(x-8)=2x(x+5)+3x(x-8)

$$( x + 5 ) ( x - 8 ) = 2 x ( x + 5 ) + 3 x ( x - 8 )$$

Whakaoti mō x (complex solution)

$$x=\frac{-\sqrt{519}i+11}{8}\approx 1.375-2.847696437i$$

$$x=\frac{11+\sqrt{519}i}{8}\approx 1.375+2.847696437i$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakamahia te āhuatanga tuaritanga hei whakarea te x+5 ki te x-8 ka whakakotahi i ngā kupu rite.

$$x^{2}-3x-40=2x\left(x+5\right)+3x\left(x-8\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 2x ki te x+5.

$$x^{2}-3x-40=2x^{2}+10x+3x\left(x-8\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 3x ki te x-8.

$$x^{2}-3x-40=2x^{2}+10x+3x^{2}-24x$$

Pahekotia te 2x^{2} me 3x^{2}, ka 5x^{2}.

$$x^{2}-3x-40=5x^{2}+10x-24x$$

Pahekotia te 10x me -24x, ka -14x.

$$x^{2}-3x-40=5x^{2}-14x$$

Tangohia te 5x^{2} mai i ngā taha e rua.

$$x^{2}-3x-40-5x^{2}=-14x$$

Pahekotia te x^{2} me -5x^{2}, ka -4x^{2}.

$$-4x^{2}-3x-40=-14x$$

Me tāpiri te 14x ki ngā taha e rua.

$$-4x^{2}-3x-40+14x=0$$

Pahekotia te -3x me 14x, ka 11x.

$$-4x^{2}+11x-40=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi -4 mō a, 11 mō b, me -40 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-11±\sqrt{11^{2}-4\left(-4\right)\left(-40\right)}}{2\left(-4\right)}$$

Pūrua 11.

$$x=\frac{-11±\sqrt{121-4\left(-4\right)\left(-40\right)}}{2\left(-4\right)}$$

Whakareatia -4 ki te -4.

$$x=\frac{-11±\sqrt{121+16\left(-40\right)}}{2\left(-4\right)}$$

Whakareatia 16 ki te -40.

$$x=\frac{-11±\sqrt{121-640}}{2\left(-4\right)}$$

Tāpiri 121 ki te -640.

$$x=\frac{-11±\sqrt{-519}}{2\left(-4\right)}$$

Tuhia te pūtakerua o te -519.

$$x=\frac{-11±\sqrt{519}i}{2\left(-4\right)}$$

Whakareatia 2 ki te -4.

$$x=\frac{-11±\sqrt{519}i}{-8}$$

Nā, me whakaoti te whārite x=\frac{-11±\sqrt{519}i}{-8} ina he tāpiri te ±. Tāpiri -11 ki te i\sqrt{519}.

$$x=\frac{-11+\sqrt{519}i}{-8}$$

Whakawehe -11+i\sqrt{519} ki te -8.

$$x=\frac{-\sqrt{519}i+11}{8}$$

Nā, me whakaoti te whārite x=\frac{-11±\sqrt{519}i}{-8} ina he tango te ±. Tango i\sqrt{519} mai i -11.

$$x=\frac{-\sqrt{519}i-11}{-8}$$

Whakawehe -11-i\sqrt{519} ki te -8.

$$x=\frac{11+\sqrt{519}i}{8}$$

Kua oti te whārite te whakatau.

$$x=\frac{-\sqrt{519}i+11}{8} x=\frac{11+\sqrt{519}i}{8}$$

Ngā Upane mō te Whakaoti i te Pūrua

Whakamahia te āhuatanga tuaritanga hei whakarea te x+5 ki te x-8 ka whakakotahi i ngā kupu rite.

$$x^{2}-3x-40=2x\left(x+5\right)+3x\left(x-8\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 2x ki te x+5.

$$x^{2}-3x-40=2x^{2}+10x+3x\left(x-8\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 3x ki te x-8.

$$x^{2}-3x-40=2x^{2}+10x+3x^{2}-24x$$

Pahekotia te 2x^{2} me 3x^{2}, ka 5x^{2}.

$$x^{2}-3x-40=5x^{2}+10x-24x$$

Pahekotia te 10x me -24x, ka -14x.

$$x^{2}-3x-40=5x^{2}-14x$$

Tangohia te 5x^{2} mai i ngā taha e rua.

$$x^{2}-3x-40-5x^{2}=-14x$$

Pahekotia te x^{2} me -5x^{2}, ka -4x^{2}.

$$-4x^{2}-3x-40=-14x$$

Me tāpiri te 14x ki ngā taha e rua.

$$-4x^{2}-3x-40+14x=0$$

Pahekotia te -3x me 14x, ka 11x.

$$-4x^{2}+11x-40=0$$

Me tāpiri te 40 ki ngā taha e rua. Ko te tau i tāpiria he kore ka hua koia tonu.

$$-4x^{2}+11x=40$$

Whakawehea ngā taha e rua ki te -4.

$$\frac{-4x^{2}+11x}{-4}=\frac{40}{-4}$$

Mā te whakawehe ki te -4 ka wetekia te whakareanga ki te -4.

$$x^{2}+\frac{11}{-4}x=\frac{40}{-4}$$

Whakawehe 11 ki te -4.

$$x^{2}-\frac{11}{4}x=\frac{40}{-4}$$

Whakawehe 40 ki te -4.

$$x^{2}-\frac{11}{4}x=-10$$

Whakawehea te -\frac{11}{4}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{11}{8}. Nā, tāpiria te pūrua o te -\frac{11}{8} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}-\frac{11}{4}x+\left(-\frac{11}{8}\right)^{2}=-10+\left(-\frac{11}{8}\right)^{2}$$

Pūruatia -\frac{11}{8} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-\frac{11}{4}x+\frac{121}{64}=-10+\frac{121}{64}$$

Tāpiri -10 ki te \frac{121}{64}.

$$x^{2}-\frac{11}{4}x+\frac{121}{64}=-\frac{519}{64}$$

Tauwehea x^{2}-\frac{11}{4}x+\frac{121}{64}. 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}.

$$\left(x-\frac{11}{8}\right)^{2}=-\frac{519}{64}$$

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

$$\sqrt{\left(x-\frac{11}{8}\right)^{2}}=\sqrt{-\frac{519}{64}}$$

Whakarūnātia.

$$x-\frac{11}{8}=\frac{\sqrt{519}i}{8} x-\frac{11}{8}=-\frac{\sqrt{519}i}{8}$$

Me tāpiri \frac{11}{8} ki ngā taha e rua o te whārite.

$$x=\frac{11+\sqrt{519}i}{8} x=\frac{-\sqrt{519}i+11}{8}$$