Whakaoti i te sqrt{25-x^2}-sqrt{15+x^2}=4

$$\sqrt { 25 - x ^ { 2 } } - \sqrt { 15 + x ^ { 2 } } = 4$$

Whakaoti mō x (complex solution)

$$x=-\sqrt{11}i\approx -0-3.31662479i$$

$$x=\sqrt{11}i\approx 3.31662479i$$

Ngā Upane Otinga

Me tango -\sqrt{15+x^{2}} mai i ngā taha e rua o te whārite.

$$\sqrt{25-x^{2}}=4+\sqrt{15+x^{2}}$$

Pūruatia ngā taha e rua o te whārite.

$$\left(\sqrt{25-x^{2}}\right)^{2}=\left(4+\sqrt{15+x^{2}}\right)^{2}$$

Tātaihia te \sqrt{25-x^{2}} mā te pū o 2, kia riro ko 25-x^{2}.

$$25-x^{2}=\left(4+\sqrt{15+x^{2}}\right)^{2}$$

Whakamahia te ture huarua \left(a+b\right)^{2}=a^{2}+2ab+b^{2} hei whakaroha \left(4+\sqrt{15+x^{2}}\right)^{2}.

$$25-x^{2}=16+8\sqrt{15+x^{2}}+\left(\sqrt{15+x^{2}}\right)^{2}$$

Tātaihia te \sqrt{15+x^{2}} mā te pū o 2, kia riro ko 15+x^{2}.

$$25-x^{2}=16+8\sqrt{15+x^{2}}+15+x^{2}$$

Tāpirihia te 16 ki te 15, ka 31.

$$25-x^{2}=31+8\sqrt{15+x^{2}}+x^{2}$$

Me tango 31+x^{2} mai i ngā taha e rua o te whārite.

$$25-x^{2}-\left(31+x^{2}\right)=8\sqrt{15+x^{2}}$$

Hei kimi i te tauaro o 31+x^{2}, kimihia te tauaro o ia taurangi.

$$25-x^{2}-31-x^{2}=8\sqrt{15+x^{2}}$$

Tangohia te 31 i te 25, ka -6.

$$-6-x^{2}-x^{2}=8\sqrt{15+x^{2}}$$

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

$$-6-2x^{2}=8\sqrt{15+x^{2}}$$

Pūruatia ngā taha e rua o te whārite.

$$\left(-6-2x^{2}\right)^{2}=\left(8\sqrt{15+x^{2}}\right)^{2}$$

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(-6-2x^{2}\right)^{2}.

$$36+24x^{2}+4\left(x^{2}\right)^{2}=\left(8\sqrt{15+x^{2}}\right)^{2}$$

Hei hiki pū ki tētahi pū anō, me whakarea ngā taupū. Me whakarea te 2 me te 2 kia riro ai te 4.

$$36+24x^{2}+4x^{4}=\left(8\sqrt{15+x^{2}}\right)^{2}$$

Whakarohaina te \left(8\sqrt{15+x^{2}}\right)^{2}.

$$36+24x^{2}+4x^{4}=8^{2}\left(\sqrt{15+x^{2}}\right)^{2}$$

Tātaihia te 8 mā te pū o 2, kia riro ko 64.

$$36+24x^{2}+4x^{4}=64\left(\sqrt{15+x^{2}}\right)^{2}$$

Tātaihia te \sqrt{15+x^{2}} mā te pū o 2, kia riro ko 15+x^{2}.

$$36+24x^{2}+4x^{4}=64\left(15+x^{2}\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 64 ki te 15+x^{2}.

$$36+24x^{2}+4x^{4}=960+64x^{2}$$

Tangohia te 960 mai i ngā taha e rua.

$$36+24x^{2}+4x^{4}-960=64x^{2}$$

Tangohia te 960 i te 36, ka -924.

$$-924+24x^{2}+4x^{4}=64x^{2}$$

Tangohia te 64x^{2} mai i ngā taha e rua.

$$-924+24x^{2}+4x^{4}-64x^{2}=0$$

Pahekotia te 24x^{2} me -64x^{2}, ka -40x^{2}.

$$-924-40x^{2}+4x^{4}=0$$

Whakakapia te t mō te x^{2}.

$$4t^{2}-40t-924=0$$

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 4 mō te a, te -40 mō te b, me te -924 mō te c i te ture pūrua.

$$t=\frac{-\left(-40\right)±\sqrt{\left(-40\right)^{2}-4\times 4\left(-924\right)}}{2\times 4}$$

Mahia ngā tātaitai.

$$t=\frac{40±128}{8}$$

Whakaotia te whārite t=\frac{40±128}{8} ina he tōrunga te ±, ina he tōraro te ±.

$$t=21 t=-11$$

I te mea ko x=t^{2}, ka riro ngā otinga mā te arotake i te x=±\sqrt{t} mō ia t.

$$x=-\sqrt{21} x=\sqrt{21} x=-\sqrt{11}i x=\sqrt{11}i$$

Whakakapia te -\sqrt{21} mō te x i te whārite \sqrt{25-x^{2}}-\sqrt{15+x^{2}}=4.

$$\sqrt{25-\left(-\sqrt{21}\right)^{2}}-\sqrt{15+\left(-\sqrt{21}\right)^{2}}=4$$

Whakarūnātia. Ko te uara x=-\sqrt{21} kāore e ngata ana ki te whārite nā te mea e rerekē ngā tohu o te taha maui me te taha katau.

$$-4=4$$

Whakakapia te \sqrt{21} mō te x i te whārite \sqrt{25-x^{2}}-\sqrt{15+x^{2}}=4.

$$\sqrt{25-\left(\sqrt{21}\right)^{2}}-\sqrt{15+\left(\sqrt{21}\right)^{2}}=4$$

Whakarūnātia. Ko te uara x=\sqrt{21} kāore e ngata ana ki te whārite nā te mea e rerekē ngā tohu o te taha maui me te taha katau.

$$-4=4$$

Whakakapia te -\sqrt{11}i mō te x i te whārite \sqrt{25-x^{2}}-\sqrt{15+x^{2}}=4.

$$\sqrt{25-\left(-\sqrt{11}i\right)^{2}}-\sqrt{15+\left(-\sqrt{11}i\right)^{2}}=4$$

Whakarūnātia. Ko te uara x=-\sqrt{11}i kua ngata te whārite.

$$4=4$$

Whakakapia te \sqrt{11}i mō te x i te whārite \sqrt{25-x^{2}}-\sqrt{15+x^{2}}=4.

$$\sqrt{25-\left(\sqrt{11}i\right)^{2}}-\sqrt{15+\left(\sqrt{11}i\right)^{2}}=4$$

Whakarūnātia. Ko te uara x=\sqrt{11}i kua ngata te whārite.

$$4=4$$

Rārangihia ngā rongoā katoa o \sqrt{25-x^{2}}=\sqrt{x^{2}+15}+4.

$$x=-\sqrt{11}i x=\sqrt{11}i$$