9d^4-840d^2+631=0 eching

$$9 d ^ { 4 } - 840 d ^ { 2 } + 631 = 0$$

d uchun yechish

$$d=-\frac{\sqrt{420-3\sqrt{18969}}}{3}\approx -0,870250386$$

$$d=\frac{\sqrt{420-3\sqrt{18969}}}{3}\approx 0,870250386$$

$$d = \frac{\sqrt{3 \sqrt{18969} + 420}}{3} \approx 9,621642147$$

$$d = -\frac{\sqrt{3 \sqrt{18969} + 420}}{3} \approx -9,621642147$$

Kvadrat tenglamalaridan foydalanish qoidalari

d^{2} uchun t ni almashtiring.

$$9t^{2}-840t+631=0$$

ax^{2}+bx+c=0 shaklidagi barcha tenglamalarni bu formula bilan yechish mumkin: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Kvadrat tenglamada a uchun 9 ni, b uchun -840 ni va c uchun 631 ni ayiring.

$$t=\frac{-\left(-840\right)±\sqrt{\left(-840\right)^{2}-4\times 9\times 631}}{2\times 9}$$

Hisoblarni amalga oshiring.

$$t=\frac{840±6\sqrt{18969}}{18}$$

t=\frac{840±6\sqrt{18969}}{18} tenglamasini ± plus va ± minus boʻlgan holatida ishlang.

$$t=\frac{\sqrt{18969}+140}{3} t=\frac{140-\sqrt{18969}}{3}$$

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

$$d=\sqrt{\frac{\sqrt{18969}+140}{3}} d=-\sqrt{\frac{\sqrt{18969}+140}{3}} d=\sqrt{\frac{140-\sqrt{18969}}{3}} d=-\sqrt{\frac{140-\sqrt{18969}}{3}}$$