-x^2-6x+8 eching

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

Omil

$$-\left(x-\left(-\sqrt{17}-3\right)\right)\left(x-\left(\sqrt{17}-3\right)\right)$$

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.

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

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.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\left(-1\right)\times 8}}{2\left(-1\right)}$$

-6 kvadratini chiqarish.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\left(-1\right)\times 8}}{2\left(-1\right)}$$

-4 ni -1 marotabaga ko'paytirish.

$$x=\frac{-\left(-6\right)±\sqrt{36+4\times 8}}{2\left(-1\right)}$$

4 ni 8 marotabaga ko'paytirish.

$$x=\frac{-\left(-6\right)±\sqrt{36+32}}{2\left(-1\right)}$$

36 ni 32 ga qo'shish.

$$x=\frac{-\left(-6\right)±\sqrt{68}}{2\left(-1\right)}$$

68 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-6\right)±2\sqrt{17}}{2\left(-1\right)}$$

-6 ning teskarisi 6 ga teng.

$$x=\frac{6±2\sqrt{17}}{2\left(-1\right)}$$

2 ni -1 marotabaga ko'paytirish.

$$x=\frac{6±2\sqrt{17}}{-2}$$

x=\frac{6±2\sqrt{17}}{-2} tenglamasini yeching, bunda ± musbat. 6 ni 2\sqrt{17} ga qo'shish.

$$x=\frac{2\sqrt{17}+6}{-2}$$

6+2\sqrt{17} ni -2 ga bo'lish.

$$x=-\left(\sqrt{17}+3\right)$$

x=\frac{6±2\sqrt{17}}{-2} tenglamasini yeching, bunda ± manfiy. 6 dan 2\sqrt{17} ni ayirish.

$$x=\frac{6-2\sqrt{17}}{-2}$$

6-2\sqrt{17} ni -2 ga bo'lish.

$$x=\sqrt{17}-3$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun -\left(3+\sqrt{17}\right) ga va x_{2} uchun -3+\sqrt{17} ga bo‘ling.

$$-x^{2}-6x+8=-\left(x-\left(-\left(\sqrt{17}+3\right)\right)\right)\left(x-\left(\sqrt{17}-3\right)\right)$$

Baholash

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