Whakaoti i te (x+1)^2-(x-1)^2+12geq0

$$( x + 1 ) ^ { 2 } - ( x - 1 ) ^ { 2 } + 12 \geq 0$$

Whakaoti mō x

$$x\geq -3$$

Ngā Upane Otinga

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

$$x^{2}+2x+1-\left(x-1\right)^{2}+12\geq 0$$

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

$$x^{2}+2x+1-\left(x^{2}-2x+1\right)+12\geq 0$$

Hei kimi i te tauaro o x^{2}-2x+1, kimihia te tauaro o ia taurangi.

$$x^{2}+2x+1-x^{2}+2x-1+12\geq 0$$

Pahekotia te x^{2} me -x^{2}, ka 0.

$$2x+1+2x-1+12\geq 0$$

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

$$4x+1-1+12\geq 0$$

Tangohia te 1 i te 1, ka 0.

$$4x+12\geq 0$$

Tangohia te 12 mai i ngā taha e rua. Ko te tau i tango i te kore ka hua ko tōna korenga.

$$4x\geq -12$$

Whakawehea ngā taha e rua ki te 4. I te mea he tōrunga te 4, kāore e huri te ahunga koreōrite.

$$x\geq \frac{-12}{4}$$

Whakawehea te -12 ki te 4, kia riro ko -3.

$$x\geq -3$$