(x^2-3x+5)+(2x^2-7x-4)= eching

$$( x ^ { 2 } - 3 x + 5 ) + ( 2 x ^ { 2 } - 7 x - 4 ) =$$

Baholash

$$3x^{2}-10x+1$$

Yechim qadamlari

3x^{2} ni olish uchun x^{2} va 2x^{2} ni birlashtirish.

$$3x^{2}-3x+5-7x-4$$

-10x ni olish uchun -3x va -7x ni birlashtirish.

$$3x^{2}-10x+5-4$$

1 olish uchun 5 dan 4 ni ayirish.

$$3x^{2}-10x+1$$

Omil

$$3\left(x-\frac{5-\sqrt{22}}{3}\right)\left(x-\frac{\sqrt{22}+5}{3}\right)$$

Kvadrat tenglamalaridan foydalanish qoidalari

3x^{2} ni olish uchun x^{2} va 2x^{2} ni birlashtirish.

$$factor(3x^{2}-3x+5-7x-4)$$

-10x ni olish uchun -3x va -7x ni birlashtirish.

$$factor(3x^{2}-10x+5-4)$$

1 olish uchun 5 dan 4 ni ayirish.

$$factor(3x^{2}-10x+1)$$

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.

$$3x^{2}-10x+1=0$$

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.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 3}}{2\times 3}$$

-10 kvadratini chiqarish.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\times 3}}{2\times 3}$$

-4 ni 3 marotabaga ko'paytirish.

$$x=\frac{-\left(-10\right)±\sqrt{100-12}}{2\times 3}$$

100 ni -12 ga qo'shish.

$$x=\frac{-\left(-10\right)±\sqrt{88}}{2\times 3}$$

88 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-10\right)±2\sqrt{22}}{2\times 3}$$

-10 ning teskarisi 10 ga teng.

$$x=\frac{10±2\sqrt{22}}{2\times 3}$$

2 ni 3 marotabaga ko'paytirish.

$$x=\frac{10±2\sqrt{22}}{6}$$

x=\frac{10±2\sqrt{22}}{6} tenglamasini yeching, bunda ± musbat. 10 ni 2\sqrt{22} ga qo'shish.

$$x=\frac{2\sqrt{22}+10}{6}$$

10+2\sqrt{22} ni 6 ga bo'lish.

$$x=\frac{\sqrt{22}+5}{3}$$

x=\frac{10±2\sqrt{22}}{6} tenglamasini yeching, bunda ± manfiy. 10 dan 2\sqrt{22} ni ayirish.

$$x=\frac{10-2\sqrt{22}}{6}$$

10-2\sqrt{22} ni 6 ga bo'lish.

$$x=\frac{5-\sqrt{22}}{3}$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun \frac{5+\sqrt{22}}{3} ga va x_{2} uchun \frac{5-\sqrt{22}}{3} ga bo‘ling.

$$3x^{2}-10x+1=3\left(x-\frac{\sqrt{22}+5}{3}\right)\left(x-\frac{5-\sqrt{22}}{3}\right)$$