9x^2-155x-500 eching

$$9 x ^ { 2 } - 155 x - 500$$

Omil

$$\left(x-20\right)\left(9x+25\right)$$

Yechim qadamlari

Ifodani guruhlash orqali faktorlang. Avvalo, ifoda 9x^{2}+ax+bx-500 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-155 ab=9\left(-500\right)=-4500$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. -4500-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,-4500 2,-2250 3,-1500 4,-1125 5,-900 6,-750 9,-500 10,-450 12,-375 15,-300 18,-250 20,-225 25,-180 30,-150 36,-125 45,-100 50,-90 60,-75$$

Har bir juftlik yigʻindisini hisoblang.

$$1-4500=-4499 2-2250=-2248 3-1500=-1497 4-1125=-1121 5-900=-895 6-750=-744 9-500=-491 10-450=-440 12-375=-363 15-300=-285 18-250=-232 20-225=-205 25-180=-155 30-150=-120 36-125=-89 45-100=-55 50-90=-40 60-75=-15$$

Yechim – -155 yigʻindisini beruvchi juftlik.

$$a=-180 b=25$$

9x^{2}-155x-500 ni \left(9x^{2}-180x\right)+\left(25x-500\right) sifatida qaytadan yozish.

$$\left(9x^{2}-180x\right)+\left(25x-500\right)$$

Birinchi guruhda 9x ni va ikkinchi guruhda 25 ni faktordan chiqaring.

$$9x\left(x-20\right)+25\left(x-20\right)$$

Distributiv funktsiyasidan foydalangan holda x-20 umumiy terminini chiqaring.

$$\left(x-20\right)\left(9x+25\right)$$

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.

$$9x^{2}-155x-500=0$$

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.

$$x=\frac{-\left(-155\right)±\sqrt{\left(-155\right)^{2}-4\times 9\left(-500\right)}}{2\times 9}$$

-155 kvadratini chiqarish.

$$x=\frac{-\left(-155\right)±\sqrt{24025-4\times 9\left(-500\right)}}{2\times 9}$$

-4 ni 9 marotabaga ko'paytirish.

$$x=\frac{-\left(-155\right)±\sqrt{24025-36\left(-500\right)}}{2\times 9}$$

-36 ni -500 marotabaga ko'paytirish.

$$x=\frac{-\left(-155\right)±\sqrt{24025+18000}}{2\times 9}$$

24025 ni 18000 ga qo'shish.

$$x=\frac{-\left(-155\right)±\sqrt{42025}}{2\times 9}$$

42025 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-155\right)±205}{2\times 9}$$

-155 ning teskarisi 155 ga teng.

$$x=\frac{155±205}{2\times 9}$$

2 ni 9 marotabaga ko'paytirish.

$$x=\frac{155±205}{18}$$

x=\frac{155±205}{18} tenglamasini yeching, bunda ± musbat. 155 ni 205 ga qo'shish.

$$x=\frac{360}{18}$$

360 ni 18 ga bo'lish.

$$x=20$$

x=\frac{155±205}{18} tenglamasini yeching, bunda ± manfiy. 155 dan 205 ni ayirish.

$$x=-\frac{50}{18}$$

\frac{-50}{18} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x=-\frac{25}{9}$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun 20 ga va x_{2} uchun -\frac{25}{9} ga bo‘ling.

$$9x^{2}-155x-500=9\left(x-20\right)\left(x-\left(-\frac{25}{9}\right)\right)$$

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

$$9x^{2}-155x-500=9\left(x-20\right)\left(x+\frac{25}{9}\right)$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{25}{9} ni x ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$9x^{2}-155x-500=9\left(x-20\right)\times \frac{9x+25}{9}$$

9 va 9 ichida eng katta umumiy 9 faktorini bekor qiling.

$$9x^{2}-155x-500=\left(x-20\right)\left(9x+25\right)$$

Baholash

$$\left(x-20\right)\left(9x+25\right)$$