Whakaoti i te 3X=sqrt{X^2+6}-4

$$3 X = \sqrt { X ^ { 2 } + 6 } - 4$$

Whakaoti mō X

$$X=-\frac{1}{2}=-0.5$$

Ngā Upane Otinga

Me tango -4 mai i ngā taha e rua o te whārite.

$$3X+4=\sqrt{X^{2}+6}$$

Pūruatia ngā taha e rua o te whārite.

$$\left(3X+4\right)^{2}=\left(\sqrt{X^{2}+6}\right)^{2}$$

Whakamahia te ture huarua \left(a+b\right)^{2}=a^{2}+2ab+b^{2} hei whakaroha \left(3X+4\right)^{2}.

$$9X^{2}+24X+16=\left(\sqrt{X^{2}+6}\right)^{2}$$

Tātaihia te \sqrt{X^{2}+6} mā te pū o 2, kia riro ko X^{2}+6.

$$9X^{2}+24X+16=X^{2}+6$$

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

$$9X^{2}+24X+16-X^{2}=6$$

Pahekotia te 9X^{2} me -X^{2}, ka 8X^{2}.

$$8X^{2}+24X+16=6$$

Tangohia te 6 mai i ngā taha e rua.

$$8X^{2}+24X+16-6=0$$

Tangohia te 6 i te 16, ka 10.

$$8X^{2}+24X+10=0$$

Whakawehea ngā taha e rua ki te 2.

$$4X^{2}+12X+5=0$$

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei 4X^{2}+aX+bX+5. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=12 ab=4\times 5=20$$

I te mea kua tōrunga te ab, he ōrite te tohu o a me b. I te mea kua tōrunga te a+b, he tōrunga hoki a a me b. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua 20.

$$1,20 2,10 4,5$$

Tātaihia te tapeke mō ia takirua.

$$1+20=21 2+10=12 4+5=9$$

Ko te otinga te takirua ka hoatu i te tapeke 12.

$$a=2 b=10$$

Tuhia anō te 4X^{2}+12X+5 hei \left(4X^{2}+2X\right)+\left(10X+5\right).

$$\left(4X^{2}+2X\right)+\left(10X+5\right)$$

Tauwehea te 2X i te tuatahi me te 5 i te rōpū tuarua.

$$2X\left(2X+1\right)+5\left(2X+1\right)$$

Whakatauwehea atu te kīanga pātahi 2X+1 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(2X+1\right)\left(2X+5\right)$$

Hei kimi otinga whārite, me whakaoti te 2X+1=0 me te 2X+5=0.

$$X=-\frac{1}{2} X=-\frac{5}{2}$$

Whakakapia te -\frac{1}{2} mō te X i te whārite 3X=\sqrt{X^{2}+6}-4.

$$3\left(-\frac{1}{2}\right)=\sqrt{\left(-\frac{1}{2}\right)^{2}+6}-4$$

Whakarūnātia. Ko te uara X=-\frac{1}{2} kua ngata te whārite.

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

Whakakapia te -\frac{5}{2} mō te X i te whārite 3X=\sqrt{X^{2}+6}-4.

$$3\left(-\frac{5}{2}\right)=\sqrt{\left(-\frac{5}{2}\right)^{2}+6}-4$$

Whakarūnātia. Ko te uara X=-\frac{5}{2} kāore e ngata ana ki te whārite.

$$-\frac{15}{2}=-\frac{1}{2}$$

Ko te whārite 3X+4=\sqrt{X^{2}+6} he rongoā ahurei.

$$X=-\frac{1}{2}$$