a^2+6a+9a^2-9 eching

$$a ^ { 2 } + 6 a + 9 a ^ { 2 } - 9$$

Omil

$$10\left(a-\frac{-3\sqrt{11}-3}{10}\right)\left(a-\frac{3\sqrt{11}-3}{10}\right)$$

Kvadrat tenglamalaridan foydalanish qoidalari

10a^{2} ni olish uchun a^{2} va 9a^{2} ni birlashtirish.

$$factor(10a^{2}+6a-9)$$

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.

$$10a^{2}+6a-9=0$$

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.

$$a=\frac{-6±\sqrt{6^{2}-4\times 10\left(-9\right)}}{2\times 10}$$

6 kvadratini chiqarish.

$$a=\frac{-6±\sqrt{36-4\times 10\left(-9\right)}}{2\times 10}$$

-4 ni 10 marotabaga ko'paytirish.

$$a=\frac{-6±\sqrt{36-40\left(-9\right)}}{2\times 10}$$

-40 ni -9 marotabaga ko'paytirish.

$$a=\frac{-6±\sqrt{36+360}}{2\times 10}$$

36 ni 360 ga qo'shish.

$$a=\frac{-6±\sqrt{396}}{2\times 10}$$

396 ning kvadrat ildizini chiqarish.

$$a=\frac{-6±6\sqrt{11}}{2\times 10}$$

2 ni 10 marotabaga ko'paytirish.

$$a=\frac{-6±6\sqrt{11}}{20}$$

a=\frac{-6±6\sqrt{11}}{20} tenglamasini yeching, bunda ± musbat. -6 ni 6\sqrt{11} ga qo'shish.

$$a=\frac{6\sqrt{11}-6}{20}$$

-6+6\sqrt{11} ni 20 ga bo'lish.

$$a=\frac{3\sqrt{11}-3}{10}$$

a=\frac{-6±6\sqrt{11}}{20} tenglamasini yeching, bunda ± manfiy. -6 dan 6\sqrt{11} ni ayirish.

$$a=\frac{-6\sqrt{11}-6}{20}$$

-6-6\sqrt{11} ni 20 ga bo'lish.

$$a=\frac{-3\sqrt{11}-3}{10}$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun \frac{-3+3\sqrt{11}}{10} ga va x_{2} uchun \frac{-3-3\sqrt{11}}{10} ga bo‘ling.

$$10a^{2}+6a-9=10\left(a-\frac{3\sqrt{11}-3}{10}\right)\left(a-\frac{-3\sqrt{11}-3}{10}\right)$$

Baholash

$$10a^{2}+6a-9$$

Yechim qadamlari

10a^{2} ni olish uchun a^{2} va 9a^{2} ni birlashtirish.

$$10a^{2}+6a-9$$