6x^2-7x-3= eching

$$6 x ^ { 2 } - 7 x - 3 =$$

Omil

$$\left(2x-3\right)\left(3x+1\right)$$

Yechim qadamlari

Ifodani guruhlash orqali faktorlang. Avvalo, ifoda 6x^{2}+ax+bx-3 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-7 ab=6\left(-3\right)=-18$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. -18-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,-18 2,-9 3,-6$$

Har bir juftlik yigʻindisini hisoblang.

$$1-18=-17 2-9=-7 3-6=-3$$

Yechim – -7 yigʻindisini beruvchi juftlik.

$$a=-9 b=2$$

6x^{2}-7x-3 ni \left(6x^{2}-9x\right)+\left(2x-3\right) sifatida qaytadan yozish.

$$\left(6x^{2}-9x\right)+\left(2x-3\right)$$

6x^{2}-9x ichida 3x ni ajrating.

$$3x\left(2x-3\right)+2x-3$$

Distributiv funktsiyasidan foydalangan holda 2x-3 umumiy terminini chiqaring.

$$\left(2x-3\right)\left(3x+1\right)$$

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.

$$6x^{2}-7x-3=0$$

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.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 6\left(-3\right)}}{2\times 6}$$

-7 kvadratini chiqarish.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\times 6\left(-3\right)}}{2\times 6}$$

-4 ni 6 marotabaga ko'paytirish.

$$x=\frac{-\left(-7\right)±\sqrt{49-24\left(-3\right)}}{2\times 6}$$

-24 ni -3 marotabaga ko'paytirish.

$$x=\frac{-\left(-7\right)±\sqrt{49+72}}{2\times 6}$$

49 ni 72 ga qo'shish.

$$x=\frac{-\left(-7\right)±\sqrt{121}}{2\times 6}$$

121 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-7\right)±11}{2\times 6}$$

-7 ning teskarisi 7 ga teng.

$$x=\frac{7±11}{2\times 6}$$

2 ni 6 marotabaga ko'paytirish.

$$x=\frac{7±11}{12}$$

x=\frac{7±11}{12} tenglamasini yeching, bunda ± musbat. 7 ni 11 ga qo'shish.

$$x=\frac{18}{12}$$

\frac{18}{12} ulushini 6 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x=\frac{3}{2}$$

x=\frac{7±11}{12} tenglamasini yeching, bunda ± manfiy. 7 dan 11 ni ayirish.

$$x=-\frac{4}{12}$$

\frac{-4}{12} ulushini 4 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x=-\frac{1}{3}$$

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

$$6x^{2}-7x-3=6\left(x-\frac{3}{2}\right)\left(x-\left(-\frac{1}{3}\right)\right)$$

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

$$6x^{2}-7x-3=6\left(x-\frac{3}{2}\right)\left(x+\frac{1}{3}\right)$$

Umumiy maxrajni topib va suratlarni ayirib \frac{3}{2} ni x dan ayirish. So'ngra imkoni boricha kasrni eng kichik shartga qisqartirish.

$$6x^{2}-7x-3=6\times \frac{2x-3}{2}\left(x+\frac{1}{3}\right)$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{1}{3} ni x ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$6x^{2}-7x-3=6\times \frac{2x-3}{2}\times \frac{3x+1}{3}$$

Raqamlash sonlarini va maxraj sonlariga ko'paytirish orqali \frac{2x-3}{2} ni \frac{3x+1}{3} ga ko'paytirish. So'ngra kasrni imkoni boricha eng kam a'zoga qisqartiring.

$$6x^{2}-7x-3=6\times \frac{\left(2x-3\right)\left(3x+1\right)}{2\times 3}$$

2 ni 3 marotabaga ko'paytirish.

$$6x^{2}-7x-3=6\times \frac{\left(2x-3\right)\left(3x+1\right)}{6}$$

6 va 6 ichida eng katta umumiy 6 faktorini bekor qiling.

$$6x^{2}-7x-3=\left(2x-3\right)\left(3x+1\right)$$

Baholash

$$\left(2x-3\right)\left(3x+1\right)$$