f(x)=-2x^2-10x+1 eching

$$f ( x ) = - 2 x ^ { 2 } - 10 x + 1$$

Omil

$$-2\left(x-\frac{-3\sqrt{3}-5}{2}\right)\left(x-\frac{3\sqrt{3}-5}{2}\right)$$

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.

$$-2x^{2}-10x+1=0$$

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.

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

-10 kvadratini chiqarish.

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

-4 ni -2 marotabaga ko'paytirish.

$$x=\frac{-\left(-10\right)±\sqrt{100+8}}{2\left(-2\right)}$$

100 ni 8 ga qo'shish.

$$x=\frac{-\left(-10\right)±\sqrt{108}}{2\left(-2\right)}$$

108 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-10\right)±6\sqrt{3}}{2\left(-2\right)}$$

-10 ning teskarisi 10 ga teng.

$$x=\frac{10±6\sqrt{3}}{2\left(-2\right)}$$

2 ni -2 marotabaga ko'paytirish.

$$x=\frac{10±6\sqrt{3}}{-4}$$

x=\frac{10±6\sqrt{3}}{-4} tenglamasini yeching, bunda ± musbat. 10 ni 6\sqrt{3} ga qo'shish.

$$x=\frac{6\sqrt{3}+10}{-4}$$

10+6\sqrt{3} ni -4 ga bo'lish.

$$x=\frac{-3\sqrt{3}-5}{2}$$

x=\frac{10±6\sqrt{3}}{-4} tenglamasini yeching, bunda ± manfiy. 10 dan 6\sqrt{3} ni ayirish.

$$x=\frac{10-6\sqrt{3}}{-4}$$

10-6\sqrt{3} ni -4 ga bo'lish.

$$x=\frac{3\sqrt{3}-5}{2}$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun \frac{-5-3\sqrt{3}}{2} ga va x_{2} uchun \frac{-5+3\sqrt{3}}{2} ga bo‘ling.

$$-2x^{2}-10x+1=-2\left(x-\frac{-3\sqrt{3}-5}{2}\right)\left(x-\frac{3\sqrt{3}-5}{2}\right)$$

Baholash

$$1-10x-2x^{2}$$