Whakaoti i te sqrt{2x}+2=1/2x

$$\sqrt { 2 x } + 2 = \frac { 1 } { 2 } x$$

Whakaoti mō x

$$x=4\left(\sqrt{3}+2\right)\approx 14.92820323$$

Ngā Upane Otinga

Me tango 2 mai i ngā taha e rua o te whārite.

$$\sqrt{2x}=\frac{1}{2}x-2$$

Pūruatia ngā taha e rua o te whārite.

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

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

$$2x=\left(\frac{1}{2}x-2\right)^{2}$$

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(\frac{1}{2}x-2\right)^{2}.

$$2x=\frac{1}{4}x^{2}-2x+4$$

Tangohia te \frac{1}{4}x^{2} mai i ngā taha e rua.

$$2x-\frac{1}{4}x^{2}=-2x+4$$

Me tāpiri te 2x ki ngā taha e rua.

$$2x-\frac{1}{4}x^{2}+2x=4$$

Pahekotia te 2x me 2x, ka 4x.

$$4x-\frac{1}{4}x^{2}=4$$

Tangohia te 4 mai i ngā taha e rua.

$$4x-\frac{1}{4}x^{2}-4=0$$

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.

$$-\frac{1}{4}x^{2}+4x-4=0$$

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

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

Pūrua 4.

$$x=\frac{-4±\sqrt{16-4\left(-\frac{1}{4}\right)\left(-4\right)}}{2\left(-\frac{1}{4}\right)}$$

Whakareatia -4 ki te -\frac{1}{4}.

$$x=\frac{-4±\sqrt{16-4}}{2\left(-\frac{1}{4}\right)}$$

Tāpiri 16 ki te -4.

$$x=\frac{-4±\sqrt{12}}{2\left(-\frac{1}{4}\right)}$$

Tuhia te pūtakerua o te 12.

$$x=\frac{-4±2\sqrt{3}}{2\left(-\frac{1}{4}\right)}$$

Whakareatia 2 ki te -\frac{1}{4}.

$$x=\frac{-4±2\sqrt{3}}{-\frac{1}{2}}$$

Nā, me whakaoti te whārite x=\frac{-4±2\sqrt{3}}{-\frac{1}{2}} ina he tāpiri te ±. Tāpiri -4 ki te 2\sqrt{3}.

$$x=\frac{2\sqrt{3}-4}{-\frac{1}{2}}$$

Whakawehe -4+2\sqrt{3} ki te -\frac{1}{2} mā te whakarea -4+2\sqrt{3} ki te tau huripoki o -\frac{1}{2}.

$$x=8-4\sqrt{3}$$

Nā, me whakaoti te whārite x=\frac{-4±2\sqrt{3}}{-\frac{1}{2}} ina he tango te ±. Tango 2\sqrt{3} mai i -4.

$$x=\frac{-2\sqrt{3}-4}{-\frac{1}{2}}$$

Whakawehe -4-2\sqrt{3} ki te -\frac{1}{2} mā te whakarea -4-2\sqrt{3} ki te tau huripoki o -\frac{1}{2}.

$$x=4\sqrt{3}+8$$

Kua oti te whārite te whakatau.

$$x=8-4\sqrt{3} x=4\sqrt{3}+8$$

Whakakapia te 8-4\sqrt{3} mō te x i te whārite \sqrt{2x}+2=\frac{1}{2}x.

$$\sqrt{2\left(8-4\sqrt{3}\right)}+2=\frac{1}{2}\left(8-4\sqrt{3}\right)$$

Whakarūnātia. Ko te uara x=8-4\sqrt{3} kāore e ngata ana ki te whārite.

$$2\times 3^{\frac{1}{2}}=4-2\times 3^{\frac{1}{2}}$$

Whakakapia te 4\sqrt{3}+8 mō te x i te whārite \sqrt{2x}+2=\frac{1}{2}x.

$$\sqrt{2\left(4\sqrt{3}+8\right)}+2=\frac{1}{2}\left(4\sqrt{3}+8\right)$$

Whakarūnātia. Ko te uara x=4\sqrt{3}+8 kua ngata te whārite.

$$2\times 3^{\frac{1}{2}}+4=2\times 3^{\frac{1}{2}}+4$$

Ko te whārite \sqrt{2x}=\frac{x}{2}-2 he rongoā ahurei.

$$x=4\sqrt{3}+8$$