x^2+8x-10 eching

$${ x }^{ 2 } +8x-10$$

Omil

$$\left(x-\left(-\sqrt{26}-4\right)\right)\left(x-\left(\sqrt{26}-4\right)\right)$$

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.

$$x^{2}+8x-10=0$$

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.

$$x=\frac{-8±\sqrt{8^{2}-4\left(-10\right)}}{2}$$

8 kvadratini chiqarish.

$$x=\frac{-8±\sqrt{64-4\left(-10\right)}}{2}$$

-4 ni -10 marotabaga ko'paytirish.

$$x=\frac{-8±\sqrt{64+40}}{2}$$

64 ni 40 ga qo'shish.

$$x=\frac{-8±\sqrt{104}}{2}$$

104 ning kvadrat ildizini chiqarish.

$$x=\frac{-8±2\sqrt{26}}{2}$$

x=\frac{-8±2\sqrt{26}}{2} tenglamasini yeching, bunda ± musbat. -8 ni 2\sqrt{26} ga qo'shish.

$$x=\frac{2\sqrt{26}-8}{2}$$

-8+2\sqrt{26} ni 2 ga bo'lish.

$$x=\sqrt{26}-4$$

x=\frac{-8±2\sqrt{26}}{2} tenglamasini yeching, bunda ± manfiy. -8 dan 2\sqrt{26} ni ayirish.

$$x=\frac{-2\sqrt{26}-8}{2}$$

-8-2\sqrt{26} ni 2 ga bo'lish.

$$x=-\sqrt{26}-4$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun -4+\sqrt{26} ga va x_{2} uchun -4-\sqrt{26} ga bo‘ling.

$$x^{2}+8x-10=\left(x-\left(\sqrt{26}-4\right)\right)\left(x-\left(-\sqrt{26}-4\right)\right)$$

Baholash

$$x^{2}+8x-10$$