8x^2-6x-4 eching

$$8 x ^ { 2 } - 6 x - 4$$

Omil

$$8\left(x-\frac{3-\sqrt{41}}{8}\right)\left(x-\frac{\sqrt{41}+3}{8}\right)$$

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.

$$8x^{2}-6x-4=0$$

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.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 8\left(-4\right)}}{2\times 8}$$

-6 kvadratini chiqarish.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\times 8\left(-4\right)}}{2\times 8}$$

-4 ni 8 marotabaga ko'paytirish.

$$x=\frac{-\left(-6\right)±\sqrt{36-32\left(-4\right)}}{2\times 8}$$

-32 ni -4 marotabaga ko'paytirish.

$$x=\frac{-\left(-6\right)±\sqrt{36+128}}{2\times 8}$$

36 ni 128 ga qo'shish.

$$x=\frac{-\left(-6\right)±\sqrt{164}}{2\times 8}$$

164 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-6\right)±2\sqrt{41}}{2\times 8}$$

-6 ning teskarisi 6 ga teng.

$$x=\frac{6±2\sqrt{41}}{2\times 8}$$

2 ni 8 marotabaga ko'paytirish.

$$x=\frac{6±2\sqrt{41}}{16}$$

x=\frac{6±2\sqrt{41}}{16} tenglamasini yeching, bunda ± musbat. 6 ni 2\sqrt{41} ga qo'shish.

$$x=\frac{2\sqrt{41}+6}{16}$$

6+2\sqrt{41} ni 16 ga bo'lish.

$$x=\frac{\sqrt{41}+3}{8}$$

x=\frac{6±2\sqrt{41}}{16} tenglamasini yeching, bunda ± manfiy. 6 dan 2\sqrt{41} ni ayirish.

$$x=\frac{6-2\sqrt{41}}{16}$$

6-2\sqrt{41} ni 16 ga bo'lish.

$$x=\frac{3-\sqrt{41}}{8}$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun \frac{3+\sqrt{41}}{8} ga va x_{2} uchun \frac{3-\sqrt{41}}{8} ga bo‘ling.

$$8x^{2}-6x-4=8\left(x-\frac{\sqrt{41}+3}{8}\right)\left(x-\frac{3-\sqrt{41}}{8}\right)$$

Baholash

$$8x^{2}-6x-4$$