4x^2+x-8 eching

$$4 x ^ { 2 } + x - 8$$

Omil

$$4\left(x-\frac{-\sqrt{129}-1}{8}\right)\left(x-\frac{\sqrt{129}-1}{8}\right)$$

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.

$$4x^{2}+x-8=0$$

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.

$$x=\frac{-1±\sqrt{1^{2}-4\times 4\left(-8\right)}}{2\times 4}$$

1 kvadratini chiqarish.

$$x=\frac{-1±\sqrt{1-4\times 4\left(-8\right)}}{2\times 4}$$

-4 ni 4 marotabaga ko'paytirish.

$$x=\frac{-1±\sqrt{1-16\left(-8\right)}}{2\times 4}$$

-16 ni -8 marotabaga ko'paytirish.

$$x=\frac{-1±\sqrt{1+128}}{2\times 4}$$

1 ni 128 ga qo'shish.

$$x=\frac{-1±\sqrt{129}}{2\times 4}$$

2 ni 4 marotabaga ko'paytirish.

$$x=\frac{-1±\sqrt{129}}{8}$$

x=\frac{-1±\sqrt{129}}{8} tenglamasini yeching, bunda ± musbat. -1 ni \sqrt{129} ga qo'shish.

$$x=\frac{\sqrt{129}-1}{8}$$

x=\frac{-1±\sqrt{129}}{8} tenglamasini yeching, bunda ± manfiy. -1 dan \sqrt{129} ni ayirish.

$$x=\frac{-\sqrt{129}-1}{8}$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun \frac{-1+\sqrt{129}}{8} ga va x_{2} uchun \frac{-1-\sqrt{129}}{8} ga bo‘ling.

$$4x^{2}+x-8=4\left(x-\frac{\sqrt{129}-1}{8}\right)\left(x-\frac{-\sqrt{129}-1}{8}\right)$$

Baholash

$$4x^{2}+x-8$$