-x^2-3x+54 eching

$$- x ^ { 2 } - 3 x + 54$$

Omil

$$\left(6-x\right)\left(x+9\right)$$

Yechim qadamlari

Ifodani guruhlash orqali faktorlang. Avvalo, ifoda -x^{2}+ax+bx+54 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-3 ab=-54=-54$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. -54-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,-54 2,-27 3,-18 6,-9$$

Har bir juftlik yigʻindisini hisoblang.

$$1-54=-53 2-27=-25 3-18=-15 6-9=-3$$

Yechim – -3 yigʻindisini beruvchi juftlik.

$$a=6 b=-9$$

-x^{2}-3x+54 ni \left(-x^{2}+6x\right)+\left(-9x+54\right) sifatida qaytadan yozish.

$$\left(-x^{2}+6x\right)+\left(-9x+54\right)$$

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

$$x\left(-x+6\right)+9\left(-x+6\right)$$

Distributiv funktsiyasidan foydalangan holda -x+6 umumiy terminini chiqaring.

$$\left(-x+6\right)\left(x+9\right)$$

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.

$$-x^{2}-3x+54=0$$

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.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-1\right)\times 54}}{2\left(-1\right)}$$

-3 kvadratini chiqarish.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-1\right)\times 54}}{2\left(-1\right)}$$

-4 ni -1 marotabaga ko'paytirish.

$$x=\frac{-\left(-3\right)±\sqrt{9+4\times 54}}{2\left(-1\right)}$$

4 ni 54 marotabaga ko'paytirish.

$$x=\frac{-\left(-3\right)±\sqrt{9+216}}{2\left(-1\right)}$$

9 ni 216 ga qo'shish.

$$x=\frac{-\left(-3\right)±\sqrt{225}}{2\left(-1\right)}$$

225 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-3\right)±15}{2\left(-1\right)}$$

-3 ning teskarisi 3 ga teng.

$$x=\frac{3±15}{2\left(-1\right)}$$

2 ni -1 marotabaga ko'paytirish.

$$x=\frac{3±15}{-2}$$

x=\frac{3±15}{-2} tenglamasini yeching, bunda ± musbat. 3 ni 15 ga qo'shish.

$$x=\frac{18}{-2}$$

18 ni -2 ga bo'lish.

$$x=-9$$

x=\frac{3±15}{-2} tenglamasini yeching, bunda ± manfiy. 3 dan 15 ni ayirish.

$$x=-\frac{12}{-2}$$

-12 ni -2 ga bo'lish.

$$x=6$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun -9 ga va x_{2} uchun 6 ga bo‘ling.

$$-x^{2}-3x+54=-\left(x-\left(-9\right)\right)\left(x-6\right)$$

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

$$-x^{2}-3x+54=-\left(x+9\right)\left(x-6\right)$$

Baholash

$$\left(6-x\right)\left(x+9\right)$$