Whakaoti i te 25+x^2-21/10x=1/2

$$\frac { 25 + x ^ { 2 } - 21 } { 10 x } = \frac { 1 } { 2 }$$

Whakaoti mō x

$$x=1$$

$$x=4$$

Ngā Upane Whakamahi Tauwehe

Tē taea kia ōrite te tāupe x ki 0 nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te 10x, arā, te tauraro pātahi he tino iti rawa te kitea o 10x,2.

$$25+x^{2}-21=5x$$

Tangohia te 21 i te 25, ka 4.

$$4+x^{2}=5x$$

Tangohia te 5x mai i ngā taha e rua.

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

Hurinahatia te pūrau ki te āhua tānga ngahuru. Whakaraupapahia ngā kīanga tau mai i te pū teitei rawa ki te mea iti rawa.

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

Hei whakaoti i te whārite, whakatauwehea te x^{2}-5x+4 mā te whakamahi i te tātai x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-5 ab=4$$

I te mea kua tōrunga te ab, he ōrite te tohu o a me b. I te mea kua tōraro te a+b, he tōraro hoki a a me b. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua 4.

$$-1,-4 -2,-2$$

Tātaihia te tapeke mō ia takirua.

$$-1-4=-5 -2-2=-4$$

Ko te otinga te takirua ka hoatu i te tapeke -5.

$$a=-4 b=-1$$

Me tuhi anō te kīanga whakatauwehe \left(x+a\right)\left(x+b\right) mā ngā uara i tātaihia.

$$\left(x-4\right)\left(x-1\right)$$

Hei kimi otinga whārite, me whakaoti te x-4=0 me te x-1=0.

$$x=4 x=1$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

Tē taea kia ōrite te tāupe x ki 0 nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te 10x, arā, te tauraro pātahi he tino iti rawa te kitea o 10x,2.

$$25+x^{2}-21=5x$$

Tangohia te 21 i te 25, ka 4.

$$4+x^{2}=5x$$

Tangohia te 5x mai i ngā taha e rua.

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

Hurinahatia te pūrau ki te āhua tānga ngahuru. Whakaraupapahia ngā kīanga tau mai i te pū teitei rawa ki te mea iti rawa.

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

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei x^{2}+ax+bx+4. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-5 ab=1\times 4=4$$

I te mea kua tōrunga te ab, he ōrite te tohu o a me b. I te mea kua tōraro te a+b, he tōraro hoki a a me b. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua 4.

$$-1,-4 -2,-2$$

Tātaihia te tapeke mō ia takirua.

$$-1-4=-5 -2-2=-4$$

Ko te otinga te takirua ka hoatu i te tapeke -5.

$$a=-4 b=-1$$

Tuhia anō te x^{2}-5x+4 hei \left(x^{2}-4x\right)+\left(-x+4\right).

$$\left(x^{2}-4x\right)+\left(-x+4\right)$$

Tauwehea te x i te tuatahi me te -1 i te rōpū tuarua.

$$x\left(x-4\right)-\left(x-4\right)$$

Whakatauwehea atu te kīanga pātahi x-4 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(x-4\right)\left(x-1\right)$$

Hei kimi otinga whārite, me whakaoti te x-4=0 me te x-1=0.

$$x=4 x=1$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tē taea kia ōrite te tāupe x ki 0 nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te 10x, arā, te tauraro pātahi he tino iti rawa te kitea o 10x,2.

$$25+x^{2}-21=5x$$

Tangohia te 21 i te 25, ka 4.

$$4+x^{2}=5x$$

Tangohia te 5x mai i ngā taha e rua.

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

Ko ngā whārite katoa o te āhua ax^{2}+bx+c=0 ka taea te whakaoti mā te whakamahi i te tikanga tātai pūrua: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. E rua ngā otinga ka puta i te tikanga tātai pūrua, ko tētahi ina he tāpiri a ±, ā, ko tētahi ina he tango.

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

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

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

Pūrua -5.

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

Whakareatia -4 ki te 4.

$$x=\frac{-\left(-5\right)±\sqrt{25-16}}{2}$$

Tāpiri 25 ki te -16.

$$x=\frac{-\left(-5\right)±\sqrt{9}}{2}$$

Tuhia te pūtakerua o te 9.

$$x=\frac{-\left(-5\right)±3}{2}$$

Ko te tauaro o -5 ko 5.

$$x=\frac{5±3}{2}$$

Nā, me whakaoti te whārite x=\frac{5±3}{2} ina he tāpiri te ±. Tāpiri 5 ki te 3.

$$x=\frac{8}{2}$$

Whakawehe 8 ki te 2.

$$x=4$$

Nā, me whakaoti te whārite x=\frac{5±3}{2} ina he tango te ±. Tango 3 mai i 5.

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

Whakawehe 2 ki te 2.

$$x=1$$

Kua oti te whārite te whakatau.

$$x=4 x=1$$

Ngā Upane mō te Whakaoti i te Pūrua

Tē taea kia ōrite te tāupe x ki 0 nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te 10x, arā, te tauraro pātahi he tino iti rawa te kitea o 10x,2.

$$25+x^{2}-21=5x$$

Tangohia te 21 i te 25, ka 4.

$$4+x^{2}=5x$$

Tangohia te 5x mai i ngā taha e rua.

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

Tangohia te 4 mai i ngā taha e rua. Ko te tau i tango i te kore ka hua ko tōna korenga.

$$x^{2}-5x=-4$$

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.

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

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

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

Tāpiri -4 ki te \frac{25}{4}.

$$x^{2}-5x+\frac{25}{4}=\frac{9}{4}$$

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}.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{9}{4}$$

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

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Whakarūnātia.

$$x-\frac{5}{2}=\frac{3}{2} x-\frac{5}{2}=-\frac{3}{2}$$

Me tāpiri \frac{5}{2} ki ngā taha e rua o te whārite.

$$x=4 x=1$$