4x^2+5x-1 eching

$$4 x ^ { 2 } + 5 x - 1$$

Omil

$$4\left(x-\frac{-\sqrt{41}-5}{8}\right)\left(x-\frac{\sqrt{41}-5}{8}\right)$$

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.

$$4x^{2}+5x-1=0$$

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.

$$x=\frac{-5±\sqrt{5^{2}-4\times 4\left(-1\right)}}{2\times 4}$$

5 kvadratini chiqarish.

$$x=\frac{-5±\sqrt{25-4\times 4\left(-1\right)}}{2\times 4}$$

-4 ni 4 marotabaga ko'paytirish.

$$x=\frac{-5±\sqrt{25-16\left(-1\right)}}{2\times 4}$$

-16 ni -1 marotabaga ko'paytirish.

$$x=\frac{-5±\sqrt{25+16}}{2\times 4}$$

25 ni 16 ga qo'shish.

$$x=\frac{-5±\sqrt{41}}{2\times 4}$$

2 ni 4 marotabaga ko'paytirish.

$$x=\frac{-5±\sqrt{41}}{8}$$

x=\frac{-5±\sqrt{41}}{8} tenglamasini yeching, bunda ± musbat. -5 ni \sqrt{41} ga qo'shish.

$$x=\frac{\sqrt{41}-5}{8}$$

x=\frac{-5±\sqrt{41}}{8} tenglamasini yeching, bunda ± manfiy. -5 dan \sqrt{41} ni ayirish.

$$x=\frac{-\sqrt{41}-5}{8}$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun \frac{-5+\sqrt{41}}{8} ga va x_{2} uchun \frac{-5-\sqrt{41}}{8} ga bo‘ling.

$$4x^{2}+5x-1=4\left(x-\frac{\sqrt{41}-5}{8}\right)\left(x-\frac{-\sqrt{41}-5}{8}\right)$$

Baholash

$$4x^{2}+5x-1$$