3d^2-3d-2 eching

$$3 d ^ { 2 } - 3 d - 2$$

Omil

$$3\left(d-\left(-\frac{\sqrt{33}}{6}+\frac{1}{2}\right)\right)\left(d-\left(\frac{\sqrt{33}}{6}+\frac{1}{2}\right)\right)$$

Kvadrat tenglamalaridan foydalanish qoidalari

Kvadrat koʻp tenglama bu orqali hisoblanadi: ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), bu yerda x_{1} va x_{2} ax^{2}+bx+c=0 kvadrat tenglamaning yechimlari.

$$3d^{2}-3d-2=0$$

ax^{2}+bx+c=0 shaklidagi barcha tenglamalarni kvadrat formulasi bilan yechish mumkin: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Kvadrat formula ikki yechmni taqdim qiladi, biri ± qo'shish bo'lganda, va ikkinchisi ayiruv bo'lganda.

$$d=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 3\left(-2\right)}}{2\times 3}$$

-3 kvadratini chiqarish.

$$d=\frac{-\left(-3\right)±\sqrt{9-4\times 3\left(-2\right)}}{2\times 3}$$

-4 ni 3 marotabaga ko'paytirish.

$$d=\frac{-\left(-3\right)±\sqrt{9-12\left(-2\right)}}{2\times 3}$$

-12 ni -2 marotabaga ko'paytirish.

$$d=\frac{-\left(-3\right)±\sqrt{9+24}}{2\times 3}$$

9 ni 24 ga qo'shish.

$$d=\frac{-\left(-3\right)±\sqrt{33}}{2\times 3}$$

-3 ning teskarisi 3 ga teng.

$$d=\frac{3±\sqrt{33}}{2\times 3}$$

2 ni 3 marotabaga ko'paytirish.

$$d=\frac{3±\sqrt{33}}{6}$$

d=\frac{3±\sqrt{33}}{6} tenglamasini yeching, bunda ± musbat. 3 ni \sqrt{33} ga qo'shish.

$$d=\frac{\sqrt{33}+3}{6}$$

3+\sqrt{33} ni 6 ga bo'lish.

$$d=\frac{\sqrt{33}}{6}+\frac{1}{2}$$

d=\frac{3±\sqrt{33}}{6} tenglamasini yeching, bunda ± manfiy. 3 dan \sqrt{33} ni ayirish.

$$d=\frac{3-\sqrt{33}}{6}$$

3-\sqrt{33} ni 6 ga bo'lish.

$$d=-\frac{\sqrt{33}}{6}+\frac{1}{2}$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun \frac{1}{2}+\frac{\sqrt{33}}{6} ga va x_{2} uchun \frac{1}{2}-\frac{\sqrt{33}}{6} ga bo‘ling.

$$3d^{2}-3d-2=3\left(d-\left(\frac{\sqrt{33}}{6}+\frac{1}{2}\right)\right)\left(d-\left(-\frac{\sqrt{33}}{6}+\frac{1}{2}\right)\right)$$

Baholash

$$3d^{2}-3d-2$$