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

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

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.

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