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

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

x uchun yechish (complex solution)

$$x=-\sqrt{11}i\approx -0-3,31662479i$$

$$x=\sqrt{11}i\approx 3,31662479i$$

Yechim qadamlari

Tenglamaning ikkala tarafidan -\sqrt{15+x^{2}} ni ayirish.

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

Tenglamaning ikkala taraf kvadratini chiqarish.

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

2 daraja ko‘rsatkichini \sqrt{25-x^{2}} ga hisoblang va 25-x^{2} ni qiymatni oling.

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

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(4+\sqrt{15+x^{2}}\right)^{2} kengaytirilishi uchun ishlating.

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

2 daraja ko‘rsatkichini \sqrt{15+x^{2}} ga hisoblang va 15+x^{2} ni qiymatni oling.

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

31 olish uchun 16 va 15'ni qo'shing.

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

Tenglamaning ikkala tarafidan 31+x^{2} ni ayirish.

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

31+x^{2} teskarisini topish uchun har birining teskarisini toping.

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

-6 olish uchun 25 dan 31 ni ayirish.

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

-2x^{2} ni olish uchun -x^{2} va -x^{2} ni birlashtirish.

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

Tenglamaning ikkala taraf kvadratini chiqarish.

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

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(-6-2x^{2}\right)^{2} kengaytirilishi uchun ishlating.

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

Daraja ko‘rsatkichini boshqa ko‘rsatkichga oshirish uchun, darajalarini ko‘paytiring. 2 va 2 ni ko‘paytirib, 4 ni oling.

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

\left(8\sqrt{15+x^{2}}\right)^{2} ni kengaytirish.

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

2 daraja ko‘rsatkichini 8 ga hisoblang va 64 ni qiymatni oling.

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

2 daraja ko‘rsatkichini \sqrt{15+x^{2}} ga hisoblang va 15+x^{2} ni qiymatni oling.

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

64 ga 15+x^{2} ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

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

Ikkala tarafdan 960 ni ayirish.

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

-924 olish uchun 36 dan 960 ni ayirish.

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

Ikkala tarafdan 64x^{2} ni ayirish.

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

-40x^{2} ni olish uchun 24x^{2} va -64x^{2} ni birlashtirish.

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

x^{2} uchun t ni almashtiring.

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

ax^{2}+bx+c=0 shaklidagi barcha tenglamalarni bu formula bilan yechish mumkin: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Kvadrat tenglamada a uchun 4 ni, b uchun -40 ni va c uchun -924 ni ayiring.

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

Hisoblarni amalga oshiring.

$$t=\frac{40±128}{8}$$

t=\frac{40±128}{8} tenglamasini ± plus va ± minus boʻlgan holatida ishlang.

$$t=21 t=-11$$

x=t^{2} boʻlganda, yechimlar har bir t uchun x=±\sqrt{t} hisoblanishi orqali olinadi.

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

\sqrt{25-x^{2}}-\sqrt{15+x^{2}}=4 tenglamasida x uchun -\sqrt{21} ni almashtiring.

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

Qisqartirish. x=-\sqrt{21} qiymati bu tenglamani qoniqtirmaydi, chunki oʻng va chap tarafdagi belgilar bir-biriga qarama-qarshi.

$$-4=4$$

\sqrt{25-x^{2}}-\sqrt{15+x^{2}}=4 tenglamasida x uchun \sqrt{21} ni almashtiring.

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

Qisqartirish. x=\sqrt{21} qiymati bu tenglamani qoniqtirmaydi, chunki oʻng va chap tarafdagi belgilar bir-biriga qarama-qarshi.

$$-4=4$$

\sqrt{25-x^{2}}-\sqrt{15+x^{2}}=4 tenglamasida x uchun -\sqrt{11}i ni almashtiring.

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

Qisqartirish. x=-\sqrt{11}i tenglamani qoniqtiradi.

$$4=4$$

\sqrt{25-x^{2}}-\sqrt{15+x^{2}}=4 tenglamasida x uchun \sqrt{11}i ni almashtiring.

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

Qisqartirish. x=\sqrt{11}i tenglamani qoniqtiradi.

$$4=4$$

\sqrt{25-x^{2}}=\sqrt{x^{2}+15}+4 boʻyicha barcha yechimlar roʻyxati.

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