-x^2+8x-13 eching

$$- x ^ { 2 } + 8 x - 13$$

Omil

$$-\left(x-\left(4-\sqrt{3}\right)\right)\left(x-\left(\sqrt{3}+4\right)\right)$$

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.

$$-x^{2}+8x-13=0$$

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.

$$x=\frac{-8±\sqrt{8^{2}-4\left(-1\right)\left(-13\right)}}{2\left(-1\right)}$$

8 kvadratini chiqarish.

$$x=\frac{-8±\sqrt{64-4\left(-1\right)\left(-13\right)}}{2\left(-1\right)}$$

-4 ni -1 marotabaga ko'paytirish.

$$x=\frac{-8±\sqrt{64+4\left(-13\right)}}{2\left(-1\right)}$$

4 ni -13 marotabaga ko'paytirish.

$$x=\frac{-8±\sqrt{64-52}}{2\left(-1\right)}$$

64 ni -52 ga qo'shish.

$$x=\frac{-8±\sqrt{12}}{2\left(-1\right)}$$

12 ning kvadrat ildizini chiqarish.

$$x=\frac{-8±2\sqrt{3}}{2\left(-1\right)}$$

2 ni -1 marotabaga ko'paytirish.

$$x=\frac{-8±2\sqrt{3}}{-2}$$

x=\frac{-8±2\sqrt{3}}{-2} tenglamasini yeching, bunda ± musbat. -8 ni 2\sqrt{3} ga qo'shish.

$$x=\frac{2\sqrt{3}-8}{-2}$$

-8+2\sqrt{3} ni -2 ga bo'lish.

$$x=4-\sqrt{3}$$

x=\frac{-8±2\sqrt{3}}{-2} tenglamasini yeching, bunda ± manfiy. -8 dan 2\sqrt{3} ni ayirish.

$$x=\frac{-2\sqrt{3}-8}{-2}$$

-8-2\sqrt{3} ni -2 ga bo'lish.

$$x=\sqrt{3}+4$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun 4-\sqrt{3} ga va x_{2} uchun 4+\sqrt{3} ga bo‘ling.

$$-x^{2}+8x-13=-\left(x-\left(4-\sqrt{3}\right)\right)\left(x-\left(\sqrt{3}+4\right)\right)$$

Baholash

$$-x^{2}+8x-13$$