(2x^2+2)^2-2(2x^2+2)-8=0 eching

$$( 2 x ^ { 2 } + 2 ) ^ { 2 } - 2 ( 2 x ^ { 2 } + 2 ) - 8 = 0$$

x uchun yechish (complex solution)

$$x=1$$

$$x=-1$$

$$x=-\sqrt{2}i\approx -0-1,414213562i$$

$$x=\sqrt{2}i\approx 1,414213562i$$

Kvadrat tenglamalaridan foydalanish qoidalari

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

$$4\left(x^{2}\right)^{2}+8x^{2}+4-2\left(2x^{2}+2\right)-8=0$$

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

$$4x^{4}+8x^{2}+4-2\left(2x^{2}+2\right)-8=0$$

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

$$4x^{4}+8x^{2}+4-4x^{2}-4-8=0$$

4x^{2} ni olish uchun 8x^{2} va -4x^{2} ni birlashtirish.

$$4x^{4}+4x^{2}+4-4-8=0$$

0 olish uchun 4 dan 4 ni ayirish.

$$4x^{4}+4x^{2}-8=0$$

x^{2} uchun t ni almashtiring.

$$4t^{2}+4t-8=0$$

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 4 ni va c uchun -8 ni ayiring.

$$t=\frac{-4±\sqrt{4^{2}-4\times 4\left(-8\right)}}{2\times 4}$$

Hisoblarni amalga oshiring.

$$t=\frac{-4±12}{8}$$

t=\frac{-4±12}{8} tenglamasini ± plus va ± minus boʻlgan holatida ishlang.

$$t=1 t=-2$$

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

$$x=-1 x=1 x=-\sqrt{2}i x=\sqrt{2}i$$

x uchun yechish

$$x=-1$$

$$x=1$$

Kvadrat tenglamalaridan foydalanish qoidalari

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

$$4\left(x^{2}\right)^{2}+8x^{2}+4-2\left(2x^{2}+2\right)-8=0$$

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

$$4x^{4}+8x^{2}+4-2\left(2x^{2}+2\right)-8=0$$

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

$$4x^{4}+8x^{2}+4-4x^{2}-4-8=0$$

4x^{2} ni olish uchun 8x^{2} va -4x^{2} ni birlashtirish.

$$4x^{4}+4x^{2}+4-4-8=0$$

0 olish uchun 4 dan 4 ni ayirish.

$$4x^{4}+4x^{2}-8=0$$

x^{2} uchun t ni almashtiring.

$$4t^{2}+4t-8=0$$

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 4 ni va c uchun -8 ni ayiring.

$$t=\frac{-4±\sqrt{4^{2}-4\times 4\left(-8\right)}}{2\times 4}$$

Hisoblarni amalga oshiring.

$$t=\frac{-4±12}{8}$$

t=\frac{-4±12}{8} tenglamasini ± plus va ± minus boʻlgan holatida ishlang.

$$t=1 t=-2$$

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

$$x=1 x=-1$$