4x^2-20x+5 eching

$$4 { x }^{ 2 } -20x+5$$

Omil

$$4\left(x-\left(\frac{5}{2}-\sqrt{5}\right)\right)\left(x-\left(\sqrt{5}+\frac{5}{2}\right)\right)$$

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.

$$4x^{2}-20x+5=0$$

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.

$$x=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times 4\times 5}}{2\times 4}$$

-20 kvadratini chiqarish.

$$x=\frac{-\left(-20\right)±\sqrt{400-4\times 4\times 5}}{2\times 4}$$

-4 ni 4 marotabaga ko'paytirish.

$$x=\frac{-\left(-20\right)±\sqrt{400-16\times 5}}{2\times 4}$$

-16 ni 5 marotabaga ko'paytirish.

$$x=\frac{-\left(-20\right)±\sqrt{400-80}}{2\times 4}$$

400 ni -80 ga qo'shish.

$$x=\frac{-\left(-20\right)±\sqrt{320}}{2\times 4}$$

320 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-20\right)±8\sqrt{5}}{2\times 4}$$

-20 ning teskarisi 20 ga teng.

$$x=\frac{20±8\sqrt{5}}{2\times 4}$$

2 ni 4 marotabaga ko'paytirish.

$$x=\frac{20±8\sqrt{5}}{8}$$

x=\frac{20±8\sqrt{5}}{8} tenglamasini yeching, bunda ± musbat. 20 ni 8\sqrt{5} ga qo'shish.

$$x=\frac{8\sqrt{5}+20}{8}$$

20+8\sqrt{5} ni 8 ga bo'lish.

$$x=\sqrt{5}+\frac{5}{2}$$

x=\frac{20±8\sqrt{5}}{8} tenglamasini yeching, bunda ± manfiy. 20 dan 8\sqrt{5} ni ayirish.

$$x=\frac{20-8\sqrt{5}}{8}$$

20-8\sqrt{5} ni 8 ga bo'lish.

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

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

$$4x^{2}-20x+5=4\left(x-\left(\sqrt{5}+\frac{5}{2}\right)\right)\left(x-\left(\frac{5}{2}-\sqrt{5}\right)\right)$$

Baholash

$$4x^{2}-20x+5$$