2x^2-18x+9 eching

$$2 { x }^{ 2 } -18x+9$$

Omil

$$2\left(x-\frac{9-3\sqrt{7}}{2}\right)\left(x-\frac{3\sqrt{7}+9}{2}\right)$$

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.

$$2x^{2}-18x+9=0$$

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.

$$x=\frac{-\left(-18\right)±\sqrt{\left(-18\right)^{2}-4\times 2\times 9}}{2\times 2}$$

-18 kvadratini chiqarish.

$$x=\frac{-\left(-18\right)±\sqrt{324-4\times 2\times 9}}{2\times 2}$$

-4 ni 2 marotabaga ko'paytirish.

$$x=\frac{-\left(-18\right)±\sqrt{324-8\times 9}}{2\times 2}$$

-8 ni 9 marotabaga ko'paytirish.

$$x=\frac{-\left(-18\right)±\sqrt{324-72}}{2\times 2}$$

324 ni -72 ga qo'shish.

$$x=\frac{-\left(-18\right)±\sqrt{252}}{2\times 2}$$

252 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-18\right)±6\sqrt{7}}{2\times 2}$$

-18 ning teskarisi 18 ga teng.

$$x=\frac{18±6\sqrt{7}}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$x=\frac{18±6\sqrt{7}}{4}$$

x=\frac{18±6\sqrt{7}}{4} tenglamasini yeching, bunda ± musbat. 18 ni 6\sqrt{7} ga qo'shish.

$$x=\frac{6\sqrt{7}+18}{4}$$

18+6\sqrt{7} ni 4 ga bo'lish.

$$x=\frac{3\sqrt{7}+9}{2}$$

x=\frac{18±6\sqrt{7}}{4} tenglamasini yeching, bunda ± manfiy. 18 dan 6\sqrt{7} ni ayirish.

$$x=\frac{18-6\sqrt{7}}{4}$$

18-6\sqrt{7} ni 4 ga bo'lish.

$$x=\frac{9-3\sqrt{7}}{2}$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun \frac{9+3\sqrt{7}}{2} ga va x_{2} uchun \frac{9-3\sqrt{7}}{2} ga bo‘ling.

$$2x^{2}-18x+9=2\left(x-\frac{3\sqrt{7}+9}{2}\right)\left(x-\frac{9-3\sqrt{7}}{2}\right)$$

Baholash

$$2x^{2}-18x+9$$