r^2-3r-130 eching

$$r ^ { 2 } - 3 r - 130$$

Omil

$$\left(r-13\right)\left(r+10\right)$$

Yechim qadamlari

Ifodani guruhlash orqali faktorlang. Avvalo, ifoda r^{2}+ar+br-130 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-3 ab=1\left(-130\right)=-130$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. -130-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,-130 2,-65 5,-26 10,-13$$

Har bir juftlik yigʻindisini hisoblang.

$$1-130=-129 2-65=-63 5-26=-21 10-13=-3$$

Yechim – -3 yigʻindisini beruvchi juftlik.

$$a=-13 b=10$$

r^{2}-3r-130 ni \left(r^{2}-13r\right)+\left(10r-130\right) sifatida qaytadan yozish.

$$\left(r^{2}-13r\right)+\left(10r-130\right)$$

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

$$r\left(r-13\right)+10\left(r-13\right)$$

Distributiv funktsiyasidan foydalangan holda r-13 umumiy terminini chiqaring.

$$\left(r-13\right)\left(r+10\right)$$

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.

$$r^{2}-3r-130=0$$

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.

$$r=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-130\right)}}{2}$$

-3 kvadratini chiqarish.

$$r=\frac{-\left(-3\right)±\sqrt{9-4\left(-130\right)}}{2}$$

-4 ni -130 marotabaga ko'paytirish.

$$r=\frac{-\left(-3\right)±\sqrt{9+520}}{2}$$

9 ni 520 ga qo'shish.

$$r=\frac{-\left(-3\right)±\sqrt{529}}{2}$$

529 ning kvadrat ildizini chiqarish.

$$r=\frac{-\left(-3\right)±23}{2}$$

-3 ning teskarisi 3 ga teng.

$$r=\frac{3±23}{2}$$

r=\frac{3±23}{2} tenglamasini yeching, bunda ± musbat. 3 ni 23 ga qo'shish.

$$r=\frac{26}{2}$$

26 ni 2 ga bo'lish.

$$r=13$$

r=\frac{3±23}{2} tenglamasini yeching, bunda ± manfiy. 3 dan 23 ni ayirish.

$$r=-\frac{20}{2}$$

-20 ni 2 ga bo'lish.

$$r=-10$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun 13 ga va x_{2} uchun -10 ga bo‘ling.

$$r^{2}-3r-130=\left(r-13\right)\left(r-\left(-10\right)\right)$$

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

$$r^{2}-3r-130=\left(r-13\right)\left(r+10\right)$$

Baholash

$$\left(r-13\right)\left(r+10\right)$$