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Whakaoti mō x
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3-x^{2}+4x\geq 0
Tangohia te 3 i te 6, ka 3.
-3+x^{2}-4x\leq 0
Me whakarea te koreōrite ki te -1 kia tōrunga ai te tau whakarea o te pū tino teitei i 3-x^{2}+4x. I te mea he tōraro a -1, ka huri te ahunga koreōrite.
-3+x^{2}-4x=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.
x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 1\left(-3\right)}}{2}
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 1 mō te a, te -4 mō te b, me te -3 mō te c i te ture pūrua.
x=\frac{4±2\sqrt{7}}{2}
Mahia ngā tātaitai.
x=\sqrt{7}+2 x=2-\sqrt{7}
Whakaotia te whārite x=\frac{4±2\sqrt{7}}{2} ina he tōrunga te ±, ina he tōraro te ±.
\left(x-\left(\sqrt{7}+2\right)\right)\left(x-\left(2-\sqrt{7}\right)\right)\leq 0
Tuhia anō te koreōrite mā te whakamahi i ngā otinga i whiwhi.
x-\left(\sqrt{7}+2\right)\geq 0 x-\left(2-\sqrt{7}\right)\leq 0
Kia ≤0 te otinga, me ≥0 rawa tētahi uara o x-\left(\sqrt{7}+2\right) me x-\left(2-\sqrt{7}\right), me ≤0 anō te uara o tētahi. Whakaarohia te tauira ina ko x-\left(\sqrt{7}+2\right)\geq 0 me x-\left(2-\sqrt{7}\right)\leq 0.
x\in \emptyset
He teka tēnei mō tētahi x ahakoa.
x-\left(2-\sqrt{7}\right)\geq 0 x-\left(\sqrt{7}+2\right)\leq 0
Whakaarohia te tauira ina ko x-\left(\sqrt{7}+2\right)\leq 0 me x-\left(2-\sqrt{7}\right)\geq 0.
x\in \begin{bmatrix}2-\sqrt{7},\sqrt{7}+2\end{bmatrix}
Te otinga e whakaea i ngā koreōrite e rua ko x\in \left[2-\sqrt{7},\sqrt{7}+2\right].
x\in \begin{bmatrix}2-\sqrt{7},\sqrt{7}+2\end{bmatrix}
Ko te otinga whakamutunga ko te whakakotahi i ngā otinga kua whiwhi.