8x^2-2x-8 eching

$$8 x ^ { 2 } - 2 x - 8$$

Omil

$$8\left(x-\frac{1-\sqrt{65}}{8}\right)\left(x-\frac{\sqrt{65}+1}{8}\right)$$

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.

$$8x^{2}-2x-8=0$$

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.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 8\left(-8\right)}}{2\times 8}$$

-2 kvadratini chiqarish.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\times 8\left(-8\right)}}{2\times 8}$$

-4 ni 8 marotabaga ko'paytirish.

$$x=\frac{-\left(-2\right)±\sqrt{4-32\left(-8\right)}}{2\times 8}$$

-32 ni -8 marotabaga ko'paytirish.

$$x=\frac{-\left(-2\right)±\sqrt{4+256}}{2\times 8}$$

4 ni 256 ga qo'shish.

$$x=\frac{-\left(-2\right)±\sqrt{260}}{2\times 8}$$

260 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-2\right)±2\sqrt{65}}{2\times 8}$$

-2 ning teskarisi 2 ga teng.

$$x=\frac{2±2\sqrt{65}}{2\times 8}$$

2 ni 8 marotabaga ko'paytirish.

$$x=\frac{2±2\sqrt{65}}{16}$$

x=\frac{2±2\sqrt{65}}{16} tenglamasini yeching, bunda ± musbat. 2 ni 2\sqrt{65} ga qo'shish.

$$x=\frac{2\sqrt{65}+2}{16}$$

2+2\sqrt{65} ni 16 ga bo'lish.

$$x=\frac{\sqrt{65}+1}{8}$$

x=\frac{2±2\sqrt{65}}{16} tenglamasini yeching, bunda ± manfiy. 2 dan 2\sqrt{65} ni ayirish.

$$x=\frac{2-2\sqrt{65}}{16}$$

2-2\sqrt{65} ni 16 ga bo'lish.

$$x=\frac{1-\sqrt{65}}{8}$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun \frac{1+\sqrt{65}}{8} ga va x_{2} uchun \frac{1-\sqrt{65}}{8} ga bo‘ling.

$$8x^{2}-2x-8=8\left(x-\frac{\sqrt{65}+1}{8}\right)\left(x-\frac{1-\sqrt{65}}{8}\right)$$

Baholash

$$8x^{2}-2x-8$$