10x^2-13x-3= eching

$$10 x ^ { 2 } - 13 x - 3 =$$

Omil

$$\left(2x-3\right)\left(5x+1\right)$$

Yechim qadamlari

Ifodani guruhlash orqali faktorlang. Avvalo, ifoda 10x^{2}+ax+bx-3 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-13 ab=10\left(-3\right)=-30$$

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 – -13 yigʻindisini beruvchi juftlik.

$$a=-15 b=2$$

10x^{2}-13x-3 ni \left(10x^{2}-15x\right)+\left(2x-3\right) sifatida qaytadan yozish.

$$\left(10x^{2}-15x\right)+\left(2x-3\right)$$

10x^{2}-15x ichida 5x ni ajrating.

$$5x\left(2x-3\right)+2x-3$$

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

$$\left(2x-3\right)\left(5x+1\right)$$

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.

$$10x^{2}-13x-3=0$$

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.

$$x=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\times 10\left(-3\right)}}{2\times 10}$$

-13 kvadratini chiqarish.

$$x=\frac{-\left(-13\right)±\sqrt{169-4\times 10\left(-3\right)}}{2\times 10}$$

-4 ni 10 marotabaga ko'paytirish.

$$x=\frac{-\left(-13\right)±\sqrt{169-40\left(-3\right)}}{2\times 10}$$

-40 ni -3 marotabaga ko'paytirish.

$$x=\frac{-\left(-13\right)±\sqrt{169+120}}{2\times 10}$$

169 ni 120 ga qo'shish.

$$x=\frac{-\left(-13\right)±\sqrt{289}}{2\times 10}$$

289 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-13\right)±17}{2\times 10}$$

-13 ning teskarisi 13 ga teng.

$$x=\frac{13±17}{2\times 10}$$

2 ni 10 marotabaga ko'paytirish.

$$x=\frac{13±17}{20}$$

x=\frac{13±17}{20} tenglamasini yeching, bunda ± musbat. 13 ni 17 ga qo'shish.

$$x=\frac{30}{20}$$

\frac{30}{20} ulushini 10 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

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

x=\frac{13±17}{20} tenglamasini yeching, bunda ± manfiy. 13 dan 17 ni ayirish.

$$x=-\frac{4}{20}$$

\frac{-4}{20} ulushini 4 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x=-\frac{1}{5}$$

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}{5} ga bo‘ling.

$$10x^{2}-13x-3=10\left(x-\frac{3}{2}\right)\left(x-\left(-\frac{1}{5}\right)\right)$$

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

$$10x^{2}-13x-3=10\left(x-\frac{3}{2}\right)\left(x+\frac{1}{5}\right)$$

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

$$10x^{2}-13x-3=10\times \frac{2x-3}{2}\left(x+\frac{1}{5}\right)$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{1}{5} ni x ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$10x^{2}-13x-3=10\times \frac{2x-3}{2}\times \frac{5x+1}{5}$$

Raqamlash sonlarini va maxraj sonlariga ko'paytirish orqali \frac{2x-3}{2} ni \frac{5x+1}{5} ga ko'paytirish. So'ngra kasrni imkoni boricha eng kam a'zoga qisqartiring.

$$10x^{2}-13x-3=10\times \frac{\left(2x-3\right)\left(5x+1\right)}{2\times 5}$$

2 ni 5 marotabaga ko'paytirish.

$$10x^{2}-13x-3=10\times \frac{\left(2x-3\right)\left(5x+1\right)}{10}$$

10 va 10 ichida eng katta umumiy 10 faktorini bekor qiling.

$$10x^{2}-13x-3=\left(2x-3\right)\left(5x+1\right)$$

Baholash

$$\left(2x-3\right)\left(5x+1\right)$$