f(x)=4x^2-17x+3 eching

$$f ( x ) = 4 x ^ { 2 } - 17 x + 3$$

Omil

$$4\left(x-\frac{17-\sqrt{241}}{8}\right)\left(x-\frac{\sqrt{241}+17}{8}\right)$$

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.

$$4x^{2}-17x+3=0$$

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.

$$x=\frac{-\left(-17\right)±\sqrt{\left(-17\right)^{2}-4\times 4\times 3}}{2\times 4}$$

-17 kvadratini chiqarish.

$$x=\frac{-\left(-17\right)±\sqrt{289-4\times 4\times 3}}{2\times 4}$$

-4 ni 4 marotabaga ko'paytirish.

$$x=\frac{-\left(-17\right)±\sqrt{289-16\times 3}}{2\times 4}$$

-16 ni 3 marotabaga ko'paytirish.

$$x=\frac{-\left(-17\right)±\sqrt{289-48}}{2\times 4}$$

289 ni -48 ga qo'shish.

$$x=\frac{-\left(-17\right)±\sqrt{241}}{2\times 4}$$

-17 ning teskarisi 17 ga teng.

$$x=\frac{17±\sqrt{241}}{2\times 4}$$

2 ni 4 marotabaga ko'paytirish.

$$x=\frac{17±\sqrt{241}}{8}$$

x=\frac{17±\sqrt{241}}{8} tenglamasini yeching, bunda ± musbat. 17 ni \sqrt{241} ga qo'shish.

$$x=\frac{\sqrt{241}+17}{8}$$

x=\frac{17±\sqrt{241}}{8} tenglamasini yeching, bunda ± manfiy. 17 dan \sqrt{241} ni ayirish.

$$x=\frac{17-\sqrt{241}}{8}$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun \frac{17+\sqrt{241}}{8} ga va x_{2} uchun \frac{17-\sqrt{241}}{8} ga bo‘ling.

$$4x^{2}-17x+3=4\left(x-\frac{\sqrt{241}+17}{8}\right)\left(x-\frac{17-\sqrt{241}}{8}\right)$$

Baholash

$$4x^{2}-17x+3$$