-2/5(x^2+1/2)^2=-3/8 eching

$$- \frac { 2 } { 5 } ( x ^ { 2 } + \frac { 1 } { 2 } ) ^ { 2 } = - \frac { 3 } { 8 }$$

x uchun yechish (complex solution)

$$x=\frac{\sqrt{\sqrt{15}-2}}{2}\approx 0,684284909$$

$$x=-\frac{\sqrt{\sqrt{15}-2}}{2}\approx -0,684284909$$

$$x=-\frac{i\sqrt{\sqrt{15}+2}}{2}\approx -0-1,211711945i$$

$$x=\frac{i\sqrt{\sqrt{15}+2}}{2}\approx 1,211711945i$$

Kvadrat tenglamalaridan foydalanish qoidalari

Ikki tarafini -\frac{5}{2} va teskari kasri -\frac{2}{5} ga ko‘paytiring.

$$\left(x^{2}+\frac{1}{2}\right)^{2}=-\frac{3}{8}\left(-\frac{5}{2}\right)$$

\frac{15}{16} hosil qilish uchun -\frac{3}{8} va -\frac{5}{2} ni ko'paytirish.

$$\left(x^{2}+\frac{1}{2}\right)^{2}=\frac{15}{16}$$

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(x^{2}+\frac{1}{2}\right)^{2} kengaytirilishi uchun ishlating.

$$\left(x^{2}\right)^{2}+x^{2}+\frac{1}{4}=\frac{15}{16}$$

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

$$x^{4}+x^{2}+\frac{1}{4}=\frac{15}{16}$$

Ikkala tarafdan \frac{15}{16} ni ayirish.

$$x^{4}+x^{2}+\frac{1}{4}-\frac{15}{16}=0$$

-\frac{11}{16} olish uchun \frac{1}{4} dan \frac{15}{16} ni ayirish.

$$x^{4}+x^{2}-\frac{11}{16}=0$$

x^{2} uchun t ni almashtiring.

$$t^{2}+t-\frac{11}{16}=0$$

ax^{2}+bx+c=0 shaklidagi barcha tenglamalarni bu formula bilan yechish mumkin: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Kvadrat tenglamada a uchun 1 ni, b uchun 1 ni va c uchun -\frac{11}{16} ni ayiring.

$$t=\frac{-1±\sqrt{1^{2}-4\times 1\left(-\frac{11}{16}\right)}}{2}$$

Hisoblarni amalga oshiring.

$$t=\frac{-1±\frac{1}{2}\sqrt{15}}{2}$$

t=\frac{-1±\frac{1}{2}\sqrt{15}}{2} tenglamasini ± plus va ± minus boʻlgan holatida ishlang.

$$t=\frac{\sqrt{15}}{4}-\frac{1}{2} t=-\frac{\sqrt{15}}{4}-\frac{1}{2}$$

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

$$x=-\frac{\sqrt{\sqrt{15}-2}}{2} x=\frac{\sqrt{\sqrt{15}-2}}{2} x=-\frac{i\sqrt{\sqrt{15}+2}}{2} x=\frac{i\sqrt{\sqrt{15}+2}}{2}$$

x uchun yechish

$$x=-\frac{\sqrt{\sqrt{15}-2}}{2}\approx -0,684284909$$

$$x=\frac{\sqrt{\sqrt{15}-2}}{2}\approx 0,684284909$$

Kvadrat tenglamalaridan foydalanish qoidalari

Ikki tarafini -\frac{5}{2} va teskari kasri -\frac{2}{5} ga ko‘paytiring.

$$\left(x^{2}+\frac{1}{2}\right)^{2}=-\frac{3}{8}\left(-\frac{5}{2}\right)$$

\frac{15}{16} hosil qilish uchun -\frac{3}{8} va -\frac{5}{2} ni ko'paytirish.

$$\left(x^{2}+\frac{1}{2}\right)^{2}=\frac{15}{16}$$

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(x^{2}+\frac{1}{2}\right)^{2} kengaytirilishi uchun ishlating.

$$\left(x^{2}\right)^{2}+x^{2}+\frac{1}{4}=\frac{15}{16}$$

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

$$x^{4}+x^{2}+\frac{1}{4}=\frac{15}{16}$$

Ikkala tarafdan \frac{15}{16} ni ayirish.

$$x^{4}+x^{2}+\frac{1}{4}-\frac{15}{16}=0$$

-\frac{11}{16} olish uchun \frac{1}{4} dan \frac{15}{16} ni ayirish.

$$x^{4}+x^{2}-\frac{11}{16}=0$$

x^{2} uchun t ni almashtiring.

$$t^{2}+t-\frac{11}{16}=0$$

ax^{2}+bx+c=0 shaklidagi barcha tenglamalarni bu formula bilan yechish mumkin: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Kvadrat tenglamada a uchun 1 ni, b uchun 1 ni va c uchun -\frac{11}{16} ni ayiring.

$$t=\frac{-1±\sqrt{1^{2}-4\times 1\left(-\frac{11}{16}\right)}}{2}$$

Hisoblarni amalga oshiring.

$$t=\frac{-1±\frac{1}{2}\sqrt{15}}{2}$$

t=\frac{-1±\frac{1}{2}\sqrt{15}}{2} tenglamasini ± plus va ± minus boʻlgan holatida ishlang.

$$t=\frac{\sqrt{15}}{4}-\frac{1}{2} t=-\frac{\sqrt{15}}{4}-\frac{1}{2}$$

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

$$x=\frac{\sqrt{\sqrt{15}-2}}{2} x=-\frac{\sqrt{\sqrt{15}-2}}{2}$$