6x^2-29x-5 eching

$$6 { x }^{ 2 } -29x-5$$

Omil

$$\left(x-5\right)\left(6x+1\right)$$

Yechim qadamlari

Ifodani guruhlash orqali faktorlang. Avvalo, ifoda 6x^{2}+ax+bx-5 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-29 ab=6\left(-5\right)=-30$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. -30-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,-30 2,-15 3,-10 5,-6$$

Har bir juftlik yigʻindisini hisoblang.

$$1-30=-29 2-15=-13 3-10=-7 5-6=-1$$

Yechim – -29 yigʻindisini beruvchi juftlik.

$$a=-30 b=1$$

6x^{2}-29x-5 ni \left(6x^{2}-30x\right)+\left(x-5\right) sifatida qaytadan yozish.

$$\left(6x^{2}-30x\right)+\left(x-5\right)$$

6x^{2}-30x ichida 6x ni ajrating.

$$6x\left(x-5\right)+x-5$$

Distributiv funktsiyasidan foydalangan holda x-5 umumiy terminini chiqaring.

$$\left(x-5\right)\left(6x+1\right)$$

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.

$$6x^{2}-29x-5=0$$

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.

$$x=\frac{-\left(-29\right)±\sqrt{\left(-29\right)^{2}-4\times 6\left(-5\right)}}{2\times 6}$$

-29 kvadratini chiqarish.

$$x=\frac{-\left(-29\right)±\sqrt{841-4\times 6\left(-5\right)}}{2\times 6}$$

-4 ni 6 marotabaga ko'paytirish.

$$x=\frac{-\left(-29\right)±\sqrt{841-24\left(-5\right)}}{2\times 6}$$

-24 ni -5 marotabaga ko'paytirish.

$$x=\frac{-\left(-29\right)±\sqrt{841+120}}{2\times 6}$$

841 ni 120 ga qo'shish.

$$x=\frac{-\left(-29\right)±\sqrt{961}}{2\times 6}$$

961 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-29\right)±31}{2\times 6}$$

-29 ning teskarisi 29 ga teng.

$$x=\frac{29±31}{2\times 6}$$

2 ni 6 marotabaga ko'paytirish.

$$x=\frac{29±31}{12}$$

x=\frac{29±31}{12} tenglamasini yeching, bunda ± musbat. 29 ni 31 ga qo'shish.

$$x=\frac{60}{12}$$

60 ni 12 ga bo'lish.

$$x=5$$

x=\frac{29±31}{12} tenglamasini yeching, bunda ± manfiy. 29 dan 31 ni ayirish.

$$x=-\frac{2}{12}$$

\frac{-2}{12} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x=-\frac{1}{6}$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun 5 ga va x_{2} uchun -\frac{1}{6} ga bo‘ling.

$$6x^{2}-29x-5=6\left(x-5\right)\left(x-\left(-\frac{1}{6}\right)\right)$$

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

$$6x^{2}-29x-5=6\left(x-5\right)\left(x+\frac{1}{6}\right)$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{1}{6} ni x ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$6x^{2}-29x-5=6\left(x-5\right)\times \frac{6x+1}{6}$$

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

$$6x^{2}-29x-5=\left(x-5\right)\left(6x+1\right)$$

Baholash

$$\left(x-5\right)\left(6x+1\right)$$