x^4+1/2x^2=17/8 eching

$$\frac { x ^ { 4 } + 1 } { 2 x ^ { 2 } } = \frac { 17 } { 8 }$$

x uchun yechish

$$x=-\frac{1}{2}=-0,5$$

$$x=\frac{1}{2}=0,5$$

$$x=2$$

$$x=-2$$

Kvadrat tenglamalaridan foydalanish qoidalari

x qiymati 0 teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini 8x^{2} ga, 2x^{2},8 ning eng kichik karralisiga ko‘paytiring.

$$4\left(x^{4}+1\right)=17x^{2}$$

4 ga x^{4}+1 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$4x^{4}+4=17x^{2}$$

Ikkala tarafdan 17x^{2} ni ayirish.

$$4x^{4}+4-17x^{2}=0$$

x^{2} uchun t ni almashtiring.

$$4t^{2}-17t+4=0$$

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 -17 ni va c uchun 4 ni ayiring.

$$t=\frac{-\left(-17\right)±\sqrt{\left(-17\right)^{2}-4\times 4\times 4}}{2\times 4}$$

Hisoblarni amalga oshiring.

$$t=\frac{17±15}{8}$$

t=\frac{17±15}{8} tenglamasini ± plus va ± minus boʻlgan holatida ishlang.

$$t=4 t=\frac{1}{4}$$

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

$$x=2 x=-2 x=\frac{1}{2} x=-\frac{1}{2}$$