x^2-8x+5 eching

$$x ^ { 2 } - 8 x + 5$$

Omil

$$\left(x-\left(4-\sqrt{11}\right)\right)\left(x-\left(\sqrt{11}+4\right)\right)$$

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.

$$x^{2}-8x+5=0$$

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.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 5}}{2}$$

-8 kvadratini chiqarish.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 5}}{2}$$

-4 ni 5 marotabaga ko'paytirish.

$$x=\frac{-\left(-8\right)±\sqrt{64-20}}{2}$$

64 ni -20 ga qo'shish.

$$x=\frac{-\left(-8\right)±\sqrt{44}}{2}$$

44 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-8\right)±2\sqrt{11}}{2}$$

-8 ning teskarisi 8 ga teng.

$$x=\frac{8±2\sqrt{11}}{2}$$

x=\frac{8±2\sqrt{11}}{2} tenglamasini yeching, bunda ± musbat. 8 ni 2\sqrt{11} ga qo'shish.

$$x=\frac{2\sqrt{11}+8}{2}$$

8+2\sqrt{11} ni 2 ga bo'lish.

$$x=\sqrt{11}+4$$

x=\frac{8±2\sqrt{11}}{2} tenglamasini yeching, bunda ± manfiy. 8 dan 2\sqrt{11} ni ayirish.

$$x=\frac{8-2\sqrt{11}}{2}$$

8-2\sqrt{11} ni 2 ga bo'lish.

$$x=4-\sqrt{11}$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun 4+\sqrt{11} ga va x_{2} uchun 4-\sqrt{11} ga bo‘ling.

$$x^{2}-8x+5=\left(x-\left(\sqrt{11}+4\right)\right)\left(x-\left(4-\sqrt{11}\right)\right)$$

Baholash

$$x^{2}-8x+5$$