Whakaoti i te x^2-4x-5=08

$${ x }^{ 2 } -4x-5 = 08$$

Whakaoti mō x

$$x=-1$$

$$x=5$$

Ngā Upane Whakamahi Tauwehe

Whakareatia te 0 ki te 8, ka 0.

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

Hei whakaoti i te whārite, whakatauwehea te x^{2}-4x-5 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.

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

I te mea kua tōraro te ab, he tauaro ngā tohu o a me b. I te mea kua tōraro te a+b, he nui ake te uara pū o te tau tōraro i tō te tōrunga. Ko te takirua anake pērā ko te otinga pūnaha.

$$a=-5 b=1$$

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

$$\left(x-5\right)\left(x+1\right)$$

Hei kimi otinga whārite, me whakaoti te x-5=0 me te x+1=0.

$$x=5 x=-1$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

Whakareatia te 0 ki te 8, ka 0.

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

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei x^{2}+ax+bx-5. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-4 ab=1\left(-5\right)=-5$$

I te mea kua tōraro te ab, he tauaro ngā tohu o a me b. I te mea kua tōraro te a+b, he nui ake te uara pū o te tau tōraro i tō te tōrunga. Ko te takirua anake pērā ko te otinga pūnaha.

$$a=-5 b=1$$

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

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

Whakatauwehea atu x i te x^{2}-5x.

$$x\left(x-5\right)+x-5$$

Whakatauwehea atu te kīanga pātahi x-5 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(x-5\right)\left(x+1\right)$$

Hei kimi otinga whārite, me whakaoti te x-5=0 me te x+1=0.

$$x=5 x=-1$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakareatia te 0 ki te 8, ka 0.

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

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

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

Pūrua -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\left(-5\right)}}{2}$$

Whakareatia -4 ki te -5.

$$x=\frac{-\left(-4\right)±\sqrt{16+20}}{2}$$

Tāpiri 16 ki te 20.

$$x=\frac{-\left(-4\right)±\sqrt{36}}{2}$$

Tuhia te pūtakerua o te 36.

$$x=\frac{-\left(-4\right)±6}{2}$$

Ko te tauaro o -4 ko 4.

$$x=\frac{4±6}{2}$$

Nā, me whakaoti te whārite x=\frac{4±6}{2} ina he tāpiri te ±. Tāpiri 4 ki te 6.

$$x=\frac{10}{2}$$

Whakawehe 10 ki te 2.

$$x=5$$

Nā, me whakaoti te whārite x=\frac{4±6}{2} ina he tango te ±. Tango 6 mai i 4.

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

Whakawehe -2 ki te 2.

$$x=-1$$

Kua oti te whārite te whakatau.

$$x=5 x=-1$$

Ngā Upane mō te Whakaoti i te Pūrua

Whakareatia te 0 ki te 8, ka 0.

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

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

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

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

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

Pūrua -2.

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

Tāpiri 5 ki te 4.

$$x^{2}-4x+4=9$$

Tauwehea x^{2}-4x+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}.

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

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

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

Whakarūnātia.

$$x-2=3 x-2=-3$$

Me tāpiri 2 ki ngā taha e rua o te whārite.

$$x=5 x=-1$$