5s^2+55s+50 eching

$$5 s ^ { 2 } + 55 s + 50$$

Omil

$$5\left(s+1\right)\left(s+10\right)$$

Yechim qadamlari

5 omili.

$$5\left(s^{2}+11s+10\right)$$

Hisoblang: s^{2}+11s+10. Ifodani guruhlash orqali faktorlang. Avvalo, ifoda s^{2}+as+bs+10 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=11 ab=1\times 10=10$$

ab musbat boʻlganda, a va b da bir xil belgi bor. a+b musbat boʻlganda, a va b ikkisi ham musbat. 10-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,10 2,5$$

Har bir juftlik yigʻindisini hisoblang.

$$1+10=11 2+5=7$$

Yechim – 11 yigʻindisini beruvchi juftlik.

$$a=1 b=10$$

s^{2}+11s+10 ni \left(s^{2}+s\right)+\left(10s+10\right) sifatida qaytadan yozish.

$$\left(s^{2}+s\right)+\left(10s+10\right)$$

Birinchi guruhda s ni va ikkinchi guruhda 10 ni faktordan chiqaring.

$$s\left(s+1\right)+10\left(s+1\right)$$

Distributiv funktsiyasidan foydalangan holda s+1 umumiy terminini chiqaring.

$$\left(s+1\right)\left(s+10\right)$$

Toʻliq ajratilgan ifodani qaytadan yozing.

$$5\left(s+1\right)\left(s+10\right)$$

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.

$$5s^{2}+55s+50=0$$

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.

$$s=\frac{-55±\sqrt{55^{2}-4\times 5\times 50}}{2\times 5}$$

55 kvadratini chiqarish.

$$s=\frac{-55±\sqrt{3025-4\times 5\times 50}}{2\times 5}$$

-4 ni 5 marotabaga ko'paytirish.

$$s=\frac{-55±\sqrt{3025-20\times 50}}{2\times 5}$$

-20 ni 50 marotabaga ko'paytirish.

$$s=\frac{-55±\sqrt{3025-1000}}{2\times 5}$$

3025 ni -1000 ga qo'shish.

$$s=\frac{-55±\sqrt{2025}}{2\times 5}$$

2025 ning kvadrat ildizini chiqarish.

$$s=\frac{-55±45}{2\times 5}$$

2 ni 5 marotabaga ko'paytirish.

$$s=\frac{-55±45}{10}$$

s=\frac{-55±45}{10} tenglamasini yeching, bunda ± musbat. -55 ni 45 ga qo'shish.

$$s=-\frac{10}{10}$$

-10 ni 10 ga bo'lish.

$$s=-1$$

s=\frac{-55±45}{10} tenglamasini yeching, bunda ± manfiy. -55 dan 45 ni ayirish.

$$s=-\frac{100}{10}$$

-100 ni 10 ga bo'lish.

$$s=-10$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun -1 ga va x_{2} uchun -10 ga bo‘ling.

$$5s^{2}+55s+50=5\left(s-\left(-1\right)\right)\left(s-\left(-10\right)\right)$$

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

$$5s^{2}+55s+50=5\left(s+1\right)\left(s+10\right)$$

Baholash

$$5\left(s+1\right)\left(s+10\right)$$