5x^2-10x-35 eching

$$5 { x }^{ 2 } -10x-35$$

Omil

$$5\left(x-\left(1-2\sqrt{2}\right)\right)\left(x-\left(2\sqrt{2}+1\right)\right)$$

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.

$$5x^{2}-10x-35=0$$

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.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 5\left(-35\right)}}{2\times 5}$$

-10 kvadratini chiqarish.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\times 5\left(-35\right)}}{2\times 5}$$

-4 ni 5 marotabaga ko'paytirish.

$$x=\frac{-\left(-10\right)±\sqrt{100-20\left(-35\right)}}{2\times 5}$$

-20 ni -35 marotabaga ko'paytirish.

$$x=\frac{-\left(-10\right)±\sqrt{100+700}}{2\times 5}$$

100 ni 700 ga qo'shish.

$$x=\frac{-\left(-10\right)±\sqrt{800}}{2\times 5}$$

800 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-10\right)±20\sqrt{2}}{2\times 5}$$

-10 ning teskarisi 10 ga teng.

$$x=\frac{10±20\sqrt{2}}{2\times 5}$$

2 ni 5 marotabaga ko'paytirish.

$$x=\frac{10±20\sqrt{2}}{10}$$

x=\frac{10±20\sqrt{2}}{10} tenglamasini yeching, bunda ± musbat. 10 ni 20\sqrt{2} ga qo'shish.

$$x=\frac{20\sqrt{2}+10}{10}$$

20\sqrt{2}+10 ni 10 ga bo'lish.

$$x=2\sqrt{2}+1$$

x=\frac{10±20\sqrt{2}}{10} tenglamasini yeching, bunda ± manfiy. 10 dan 20\sqrt{2} ni ayirish.

$$x=\frac{10-20\sqrt{2}}{10}$$

10-20\sqrt{2} ni 10 ga bo'lish.

$$x=1-2\sqrt{2}$$

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

$$5x^{2}-10x-35=5\left(x-\left(2\sqrt{2}+1\right)\right)\left(x-\left(1-2\sqrt{2}\right)\right)$$

Baholash

$$5\left(x^{2}-2x-7\right)$$