5x^2+13x+6 eching

$$5 { x }^{ 2 } +13x+6$$

Omil

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

Yechim qadamlari

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

$$a+b=13 ab=5\times 6=30$$

ab musbat boʻlganda, a va b da bir xil belgi bor. a+b musbat boʻlganda, a va b ikkisi ham musbat. 30-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

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

Har bir juftlik yigʻindisini hisoblang.

$$1+30=31 2+15=17 3+10=13 5+6=11$$

Yechim – 13 yigʻindisini beruvchi juftlik.

$$a=3 b=10$$

5x^{2}+13x+6 ni \left(5x^{2}+3x\right)+\left(10x+6\right) sifatida qaytadan yozish.

$$\left(5x^{2}+3x\right)+\left(10x+6\right)$$

Birinchi guruhda x ni va ikkinchi guruhda 2 ni faktordan chiqaring.

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

Distributiv funktsiyasidan foydalangan holda 5x+3 umumiy terminini chiqaring.

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

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.

$$5x^{2}+13x+6=0$$

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.

$$x=\frac{-13±\sqrt{13^{2}-4\times 5\times 6}}{2\times 5}$$

13 kvadratini chiqarish.

$$x=\frac{-13±\sqrt{169-4\times 5\times 6}}{2\times 5}$$

-4 ni 5 marotabaga ko'paytirish.

$$x=\frac{-13±\sqrt{169-20\times 6}}{2\times 5}$$

-20 ni 6 marotabaga ko'paytirish.

$$x=\frac{-13±\sqrt{169-120}}{2\times 5}$$

169 ni -120 ga qo'shish.

$$x=\frac{-13±\sqrt{49}}{2\times 5}$$

49 ning kvadrat ildizini chiqarish.

$$x=\frac{-13±7}{2\times 5}$$

2 ni 5 marotabaga ko'paytirish.

$$x=\frac{-13±7}{10}$$

x=\frac{-13±7}{10} tenglamasini yeching, bunda ± musbat. -13 ni 7 ga qo'shish.

$$x=-\frac{6}{10}$$

\frac{-6}{10} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x=-\frac{3}{5}$$

x=\frac{-13±7}{10} tenglamasini yeching, bunda ± manfiy. -13 dan 7 ni ayirish.

$$x=-\frac{20}{10}$$

-20 ni 10 ga bo'lish.

$$x=-2$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun -\frac{3}{5} ga va x_{2} uchun -2 ga bo‘ling.

$$5x^{2}+13x+6=5\left(x-\left(-\frac{3}{5}\right)\right)\left(x-\left(-2\right)\right)$$

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

$$5x^{2}+13x+6=5\left(x+\frac{3}{5}\right)\left(x+2\right)$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{3}{5} ni x ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$5x^{2}+13x+6=5\times \frac{5x+3}{5}\left(x+2\right)$$

5 va 5 ichida eng katta umumiy 5 faktorini bekor qiling.

$$5x^{2}+13x+6=\left(5x+3\right)\left(x+2\right)$$

Baholash

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