Réitigh sqrt{25-x^2}-sqrt{15+x^2}=4

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

Réitigh do x. (complex solution)

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

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

Céimeanna Réitigh

Bain -\sqrt{15+x^{2}} ón dá thaobh den chothromóid.

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

Cearnaigh an dá thaobh den chothromóid.

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

Ríomh cumhacht \sqrt{25-x^{2}} de 2 agus faigh 25-x^{2}.

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

Úsáid an teoirim dhéthéarmach \left(a+b\right)^{2}=a^{2}+2ab+b^{2} chun \left(4+\sqrt{15+x^{2}}\right)^{2} a leathnú.

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

Ríomh cumhacht \sqrt{15+x^{2}} de 2 agus faigh 15+x^{2}.

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

Suimigh 16 agus 15 chun 31 a fháil.

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

Bain 31+x^{2} ón dá thaobh den chothromóid.

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

Chun an mhalairt ar 31+x^{2} a aimsiú, aimsigh an mhalairt ar gach téarma.

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

Dealaigh 31 ó 25 chun -6 a fháil.

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

Comhcheangail -x^{2} agus -x^{2} chun -2x^{2} a fháil.

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

Cearnaigh an dá thaobh den chothromóid.

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

Úsáid an teoirim dhéthéarmach \left(a-b\right)^{2}=a^{2}-2ab+b^{2} chun \left(-6-2x^{2}\right)^{2} a leathnú.

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

Chun cumhacht a ardú go cumhacht eile, méadaigh na heaspónaint. Iolraigh 2 agus 2 chun 4 a bhaint amach.

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

Fairsingigh \left(8\sqrt{15+x^{2}}\right)^{2}

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

Ríomh cumhacht 8 de 2 agus faigh 64.

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

Ríomh cumhacht \sqrt{15+x^{2}} de 2 agus faigh 15+x^{2}.

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

Úsáid an t-airí dáileach chun 64 a mhéadú faoi 15+x^{2}.

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

Bain 960 ón dá thaobh.

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

Dealaigh 960 ó 36 chun -924 a fháil.

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

Bain 64x^{2} ón dá thaobh.

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

Comhcheangail 24x^{2} agus -64x^{2} chun -40x^{2} a fháil.

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

Cuir t in ionad x^{2}.

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

Is féidir gach cothromóid i bhfoirm ax^{2}+bx+c=0 a réiteach ach an fhoirmle chearnach seo a úsáid: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Cuir 4 in ionad a, -40 in ionad b agus -924 in ionad c san fhoirmle chearnach.

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

Déan áirimh.

$$t=\frac{40±128}{8}$$

Réitigh an chothromóid t=\frac{40±128}{8} nuair is ionann ± agus luach deimhneach agus ± agus luach diúltach.

$$t=21 t=-11$$

Más x=t^{2}, is féidir teacht ar na réitigh ach x=±\sqrt{t} a mheas i gcomhair gach t.

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

Cuir -\sqrt{21} in ionad x sa chothromóid \sqrt{25-x^{2}}-\sqrt{15+x^{2}}=4.

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

Simpligh. Níl an chothromóid comhlíonann luach x=-\sqrt{21} toisc nach bhfuil ag an taobh clé agus an taobh deas comharthaí os a chomhair.

$$-4=4$$

Cuir \sqrt{21} in ionad x sa chothromóid \sqrt{25-x^{2}}-\sqrt{15+x^{2}}=4.

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

Simpligh. Níl an chothromóid comhlíonann luach x=\sqrt{21} toisc nach bhfuil ag an taobh clé agus an taobh deas comharthaí os a chomhair.

$$-4=4$$

Cuir -\sqrt{11}i in ionad x sa chothromóid \sqrt{25-x^{2}}-\sqrt{15+x^{2}}=4.

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

Simpligh. An luach x=-\sqrt{11}i shásaíonn an gcothromóid.

$$4=4$$

Cuir \sqrt{11}i in ionad x sa chothromóid \sqrt{25-x^{2}}-\sqrt{15+x^{2}}=4.

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

Simpligh. An luach x=\sqrt{11}i shásaíonn an gcothromóid.

$$4=4$$

Liostaigh gach réitigh de \sqrt{25-x^{2}}=\sqrt{x^{2}+15}+4.

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