Whakaoti i te (3x-5)(x+2)leqx^2

$$( 3 x - 5 ) ( x + 2 ) \leq x ^ { 2 }$$

Whakaoti mō x

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

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakamahia te āhuatanga tuaritanga hei whakarea te 3x-5 ki te x+2 ka whakakotahi i ngā kupu rite.

$$3x^{2}+x-10\leq x^{2}$$

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

$$3x^{2}+x-10-x^{2}\leq 0$$

Pahekotia te 3x^{2} me -x^{2}, ka 2x^{2}.

$$2x^{2}+x-10\leq 0$$

Kia whakaotia te koreōrite, me tauwehe te taha mauī. Ka taea te huamaha pūrua te tauwehe mā te whakamahi i te huringa ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), ina ko x_{1} me x_{2} ngā otinga o te whārite pūrua ax^{2}+bx+c=0.

$$2x^{2}+x-10=0$$

Ka taea ngā whārite katoa o te momo ax^{2}+bx+c=0 te whakaoti mā te ture pūrua: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Whakakapia te 2 mō te a, te 1 mō te b, me te -10 mō te c i te ture pūrua.

$$x=\frac{-1±\sqrt{1^{2}-4\times 2\left(-10\right)}}{2\times 2}$$

Mahia ngā tātaitai.

$$x=\frac{-1±9}{4}$$

Whakaotia te whārite x=\frac{-1±9}{4} ina he tōrunga te ±, ina he tōraro te ±.

$$x=2 x=-\frac{5}{2}$$

Tuhia anō te koreōrite mā te whakamahi i ngā otinga i whiwhi.

$$2\left(x-2\right)\left(x+\frac{5}{2}\right)\leq 0$$

Kia ≤0 te otinga, me ≥0 rawa tētahi uara o x-2 me x+\frac{5}{2}, me ≤0 anō te uara o tētahi. Consider the case when x-2\geq 0 and x+\frac{5}{2}\leq 0.

$$x-2\geq 0 x+\frac{5}{2}\leq 0$$

He teka tēnei mō tētahi x ahakoa.

$$x\in \emptyset$$

Consider the case when x-2\leq 0 and x+\frac{5}{2}\geq 0.

$$x+\frac{5}{2}\geq 0 x-2\leq 0$$

Te otinga e whakaea i ngā koreōrite e rua ko x\in \left[-\frac{5}{2},2\right].

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

Ko te otinga whakamutunga ko te whakakotahi i ngā otinga kua whiwhi.

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