Whakaoti i te 36-4*(a^2+1)5geq0

$$36 - 4 \cdot ( a ^ { 2 } + 1 ) 5 \geq 0$$

Whakaoti mō a

$$a\in \begin{bmatrix}-\frac{2\sqrt{5}}{5},\frac{2\sqrt{5}}{5}\end{bmatrix}$$

Ngā Upane Otinga

Whakareatia te 4 ki te 5, ka 20.

$$36-20\left(a^{2}+1\right)\geq 0$$

Whakamahia te āhuatanga tohatoha hei whakarea te -20 ki te a^{2}+1.

$$36-20a^{2}-20\geq 0$$

Tangohia te 20 i te 36, ka 16.

$$16-20a^{2}\geq 0$$

Me whakarea te koreōrite ki te -1 kia tōrunga ai te tau whakarea o te pū tino teitei i 16-20a^{2}. I te mea he tōraro a -1, ka huri te ahunga koreōrite.

$$-16+20a^{2}\leq 0$$

Me tāpiri te \frac{4}{5} ki ngā taha e rua.

$$a^{2}\leq \frac{4}{5}$$

Tātaitia te pūtakerua o \frac{4}{5} kia tae ki \frac{2\sqrt{5}}{5}. Tuhia anō te \frac{4}{5} hei \left(\frac{2\sqrt{5}}{5}\right)^{2}.

$$a^{2}\leq \left(\frac{2\sqrt{5}}{5}\right)^{2}$$

E mau ana te koreōrite mō |a|\leq \frac{2\sqrt{5}}{5}.

$$|a|\leq \frac{2\sqrt{5}}{5}$$

Tuhia anō te |a|\leq \frac{2\sqrt{5}}{5} hei a\in \left[-\frac{2\sqrt{5}}{5},\frac{2\sqrt{5}}{5}\right].

$$a\in \begin{bmatrix}-\frac{2\sqrt{5}}{5},\frac{2\sqrt{5}}{5}\end{bmatrix}$$