-3x^2+6x+9= eching

$$- 3 x ^ { 2 } + 6 x + 9 =$$

Omil

$$-3\left(x-3\right)\left(x+1\right)$$

Yechim qadamlari

3 omili.

$$3\left(-x^{2}+2x+3\right)$$

Hisoblang: -x^{2}+2x+3. Ifodani guruhlash orqali faktorlang. Avvalo, ifoda -x^{2}+ax+bx+3 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=2 ab=-3=-3$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b musbat boʻlganda, musbat sonda manfiyga nisbatdan kattaroq mutlaq qiymat bor. Faqat bundan juftlik tizim yechimidir.

$$a=3 b=-1$$

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

$$\left(-x^{2}+3x\right)+\left(-x+3\right)$$

Birinchi guruhda -x ni va ikkinchi guruhda -1 ni faktordan chiqaring.

$$-x\left(x-3\right)-\left(x-3\right)$$

Distributiv funktsiyasidan foydalangan holda x-3 umumiy terminini chiqaring.

$$\left(x-3\right)\left(-x-1\right)$$

Toʻliq ajratilgan ifodani qaytadan yozing.

$$3\left(x-3\right)\left(-x-1\right)$$

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.

$$-3x^{2}+6x+9=0$$

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.

$$x=\frac{-6±\sqrt{6^{2}-4\left(-3\right)\times 9}}{2\left(-3\right)}$$

6 kvadratini chiqarish.

$$x=\frac{-6±\sqrt{36-4\left(-3\right)\times 9}}{2\left(-3\right)}$$

-4 ni -3 marotabaga ko'paytirish.

$$x=\frac{-6±\sqrt{36+12\times 9}}{2\left(-3\right)}$$

12 ni 9 marotabaga ko'paytirish.

$$x=\frac{-6±\sqrt{36+108}}{2\left(-3\right)}$$

36 ni 108 ga qo'shish.

$$x=\frac{-6±\sqrt{144}}{2\left(-3\right)}$$

144 ning kvadrat ildizini chiqarish.

$$x=\frac{-6±12}{2\left(-3\right)}$$

2 ni -3 marotabaga ko'paytirish.

$$x=\frac{-6±12}{-6}$$

x=\frac{-6±12}{-6} tenglamasini yeching, bunda ± musbat. -6 ni 12 ga qo'shish.

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

6 ni -6 ga bo'lish.

$$x=-1$$

x=\frac{-6±12}{-6} tenglamasini yeching, bunda ± manfiy. -6 dan 12 ni ayirish.

$$x=-\frac{18}{-6}$$

-18 ni -6 ga bo'lish.

$$x=3$$

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 3 ga bo‘ling.

$$-3x^{2}+6x+9=-3\left(x-\left(-1\right)\right)\left(x-3\right)$$

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

$$-3x^{2}+6x+9=-3\left(x+1\right)\left(x-3\right)$$

Baholash

$$-3\left(x-3\right)\left(x+1\right)$$