4x^2-11x-3 eching

$$4 { x }^{ 2 } -11x-3$$

Omil

$$\left(x-3\right)\left(4x+1\right)$$

Yechim qadamlari

Ifodani guruhlash orqali faktorlang. Avvalo, ifoda 4x^{2}+ax+bx-3 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-11 ab=4\left(-3\right)=-12$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. -12-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,-12 2,-6 3,-4$$

Har bir juftlik yigʻindisini hisoblang.

$$1-12=-11 2-6=-4 3-4=-1$$

Yechim – -11 yigʻindisini beruvchi juftlik.

$$a=-12 b=1$$

4x^{2}-11x-3 ni \left(4x^{2}-12x\right)+\left(x-3\right) sifatida qaytadan yozish.

$$\left(4x^{2}-12x\right)+\left(x-3\right)$$

4x^{2}-12x ichida 4x ni ajrating.

$$4x\left(x-3\right)+x-3$$

Distributiv funktsiyasidan foydalangan holda x-3 umumiy terminini chiqaring.

$$\left(x-3\right)\left(4x+1\right)$$

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.

$$4x^{2}-11x-3=0$$

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.

$$x=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 4\left(-3\right)}}{2\times 4}$$

-11 kvadratini chiqarish.

$$x=\frac{-\left(-11\right)±\sqrt{121-4\times 4\left(-3\right)}}{2\times 4}$$

-4 ni 4 marotabaga ko'paytirish.

$$x=\frac{-\left(-11\right)±\sqrt{121-16\left(-3\right)}}{2\times 4}$$

-16 ni -3 marotabaga ko'paytirish.

$$x=\frac{-\left(-11\right)±\sqrt{121+48}}{2\times 4}$$

121 ni 48 ga qo'shish.

$$x=\frac{-\left(-11\right)±\sqrt{169}}{2\times 4}$$

169 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-11\right)±13}{2\times 4}$$

-11 ning teskarisi 11 ga teng.

$$x=\frac{11±13}{2\times 4}$$

2 ni 4 marotabaga ko'paytirish.

$$x=\frac{11±13}{8}$$

x=\frac{11±13}{8} tenglamasini yeching, bunda ± musbat. 11 ni 13 ga qo'shish.

$$x=\frac{24}{8}$$

24 ni 8 ga bo'lish.

$$x=3$$

x=\frac{11±13}{8} tenglamasini yeching, bunda ± manfiy. 11 dan 13 ni ayirish.

$$x=-\frac{2}{8}$$

\frac{-2}{8} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x=-\frac{1}{4}$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun 3 ga va x_{2} uchun -\frac{1}{4} ga bo‘ling.

$$4x^{2}-11x-3=4\left(x-3\right)\left(x-\left(-\frac{1}{4}\right)\right)$$

p-\left(-q\right) shaklining barcha amallarigani p+q ga soddalashtiring.

$$4x^{2}-11x-3=4\left(x-3\right)\left(x+\frac{1}{4}\right)$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{1}{4} ni x ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$4x^{2}-11x-3=4\left(x-3\right)\times \frac{4x+1}{4}$$

4 va 4 ichida eng katta umumiy 4 faktorini bekor qiling.

$$4x^{2}-11x-3=\left(x-3\right)\left(4x+1\right)$$

Baholash

$$\left(x-3\right)\left(4x+1\right)$$