Whakaoti i te sqrt{x+5}=x+4

$$\sqrt{ x+5 } = x+4$$

Whakaoti mō x

$$x=\frac{\sqrt{5}-7}{2}\approx -2.381966011$$

Ngā Upane Otinga

Pūruatia ngā taha e rua o te whārite.

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

Tātaihia te \sqrt{x+5} mā te pū o 2, kia riro ko x+5.

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

Whakamahia te ture huarua \left(a+b\right)^{2}=a^{2}+2ab+b^{2} hei whakaroha \left(x+4\right)^{2}.

$$x+5=x^{2}+8x+16$$

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

$$x+5-x^{2}=8x+16$$

Tangohia te 8x mai i ngā taha e rua.

$$x+5-x^{2}-8x=16$$

Pahekotia te x me -8x, ka -7x.

$$-7x+5-x^{2}=16$$

Tangohia te 16 mai i ngā taha e rua.

$$-7x+5-x^{2}-16=0$$

Tangohia te 16 i te 5, ka -11.

$$-7x-11-x^{2}=0$$

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.

$$-x^{2}-7x-11=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi -1 mō a, -7 mō b, me -11 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\left(-1\right)\left(-11\right)}}{2\left(-1\right)}$$

Pūrua -7.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\left(-1\right)\left(-11\right)}}{2\left(-1\right)}$$

Whakareatia -4 ki te -1.

$$x=\frac{-\left(-7\right)±\sqrt{49+4\left(-11\right)}}{2\left(-1\right)}$$

Whakareatia 4 ki te -11.

$$x=\frac{-\left(-7\right)±\sqrt{49-44}}{2\left(-1\right)}$$

Tāpiri 49 ki te -44.

$$x=\frac{-\left(-7\right)±\sqrt{5}}{2\left(-1\right)}$$

Ko te tauaro o -7 ko 7.

$$x=\frac{7±\sqrt{5}}{2\left(-1\right)}$$

Whakareatia 2 ki te -1.

$$x=\frac{7±\sqrt{5}}{-2}$$

Nā, me whakaoti te whārite x=\frac{7±\sqrt{5}}{-2} ina he tāpiri te ±. Tāpiri 7 ki te \sqrt{5}.

$$x=\frac{\sqrt{5}+7}{-2}$$

Whakawehe 7+\sqrt{5} ki te -2.

$$x=\frac{-\sqrt{5}-7}{2}$$

Nā, me whakaoti te whārite x=\frac{7±\sqrt{5}}{-2} ina he tango te ±. Tango \sqrt{5} mai i 7.

$$x=\frac{7-\sqrt{5}}{-2}$$

Whakawehe 7-\sqrt{5} ki te -2.

$$x=\frac{\sqrt{5}-7}{2}$$

Kua oti te whārite te whakatau.

$$x=\frac{-\sqrt{5}-7}{2} x=\frac{\sqrt{5}-7}{2}$$

Whakakapia te \frac{-\sqrt{5}-7}{2} mō te x i te whārite \sqrt{x+5}=x+4.

$$\sqrt{\frac{-\sqrt{5}-7}{2}+5}=\frac{-\sqrt{5}-7}{2}+4$$

Whakarūnātia. Ko te uara x=\frac{-\sqrt{5}-7}{2} kāore e ngata ana ki te whārite nā te mea e rerekē ngā tohu o te taha maui me te taha katau.

$$-\left(\frac{1}{2}-\frac{1}{2}\times 5^{\frac{1}{2}}\right)=-\frac{1}{2}\times 5^{\frac{1}{2}}+\frac{1}{2}$$

Whakakapia te \frac{\sqrt{5}-7}{2} mō te x i te whārite \sqrt{x+5}=x+4.

$$\sqrt{\frac{\sqrt{5}-7}{2}+5}=\frac{\sqrt{5}-7}{2}+4$$

Whakarūnātia. Ko te uara x=\frac{\sqrt{5}-7}{2} kua ngata te whārite.

$$\frac{1}{2}+\frac{1}{2}\times 5^{\frac{1}{2}}=\frac{1}{2}\times 5^{\frac{1}{2}}+\frac{1}{2}$$

Ko te whārite \sqrt{x+5}=x+4 he rongoā ahurei.

$$x=\frac{\sqrt{5}-7}{2}$$