Whakaoti i te x+1=sqrt{2x+5}

$$x + 1 = \sqrt { 2 x + 5 }$$

Whakaoti mō x

$$x=2$$

Ngā Upane Otinga

Pūruatia ngā taha e rua o te whārite.

$$\left(x+1\right)^{2}=\left(\sqrt{2x+5}\right)^{2}$$

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

$$x^{2}+2x+1=\left(\sqrt{2x+5}\right)^{2}$$

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

$$x^{2}+2x+1=2x+5$$

Tangohia te 2x mai i ngā taha e rua.

$$x^{2}+2x+1-2x=5$$

Pahekotia te 2x me -2x, ka 0.

$$x^{2}+1=5$$

Tangohia te 5 mai i ngā taha e rua.

$$x^{2}+1-5=0$$

Tangohia te 5 i te 1, ka -4.

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

Whakaarohia te x^{2}-4. Tuhia anō te x^{2}-4 hei x^{2}-2^{2}. Ka taea te rerekētanga o ngā pūrua te whakatauwehe mā te ture: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(x-2\right)\left(x+2\right)=0$$

Hei kimi otinga whārite, me whakaoti te x-2=0 me te x+2=0.

$$x=2 x=-2$$

Whakakapia te 2 mō te x i te whārite x+1=\sqrt{2x+5}.

$$2+1=\sqrt{2\times 2+5}$$

Whakarūnātia. Ko te uara x=2 kua ngata te whārite.

$$3=3$$

Whakakapia te -2 mō te x i te whārite x+1=\sqrt{2x+5}.

$$-2+1=\sqrt{2\left(-2\right)+5}$$

Whakarūnātia. Ko te uara x=-2 kāore e ngata ana ki te whārite nā te mea e rerekē ngā tohu o te taha maui me te taha katau.

$$-1=1$$

Ko te whārite x+1=\sqrt{2x+5} he rongoā ahurei.

$$x=2$$