3x^2-9x+1 eching

$$3 { x }^{ 2 } -9x+1$$

Omil

$$3\left(x-\left(-\frac{\sqrt{69}}{6}+\frac{3}{2}\right)\right)\left(x-\left(\frac{\sqrt{69}}{6}+\frac{3}{2}\right)\right)$$

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.

$$3x^{2}-9x+1=0$$

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.

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

-9 kvadratini chiqarish.

$$x=\frac{-\left(-9\right)±\sqrt{81-4\times 3}}{2\times 3}$$

-4 ni 3 marotabaga ko'paytirish.

$$x=\frac{-\left(-9\right)±\sqrt{81-12}}{2\times 3}$$

81 ni -12 ga qo'shish.

$$x=\frac{-\left(-9\right)±\sqrt{69}}{2\times 3}$$

-9 ning teskarisi 9 ga teng.

$$x=\frac{9±\sqrt{69}}{2\times 3}$$

2 ni 3 marotabaga ko'paytirish.

$$x=\frac{9±\sqrt{69}}{6}$$

x=\frac{9±\sqrt{69}}{6} tenglamasini yeching, bunda ± musbat. 9 ni \sqrt{69} ga qo'shish.

$$x=\frac{\sqrt{69}+9}{6}$$

9+\sqrt{69} ni 6 ga bo'lish.

$$x=\frac{\sqrt{69}}{6}+\frac{3}{2}$$

x=\frac{9±\sqrt{69}}{6} tenglamasini yeching, bunda ± manfiy. 9 dan \sqrt{69} ni ayirish.

$$x=\frac{9-\sqrt{69}}{6}$$

9-\sqrt{69} ni 6 ga bo'lish.

$$x=-\frac{\sqrt{69}}{6}+\frac{3}{2}$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun \frac{3}{2}+\frac{\sqrt{69}}{6} ga va x_{2} uchun \frac{3}{2}-\frac{\sqrt{69}}{6} ga bo‘ling.

$$3x^{2}-9x+1=3\left(x-\left(\frac{\sqrt{69}}{6}+\frac{3}{2}\right)\right)\left(x-\left(-\frac{\sqrt{69}}{6}+\frac{3}{2}\right)\right)$$

Baholash

$$3x^{2}-9x+1$$