x^2-42x+45= eching

$${ x }^{ 2 } -42x+45=$$

Omil

$$\left(x-\left(21-6\sqrt{11}\right)\right)\left(x-\left(6\sqrt{11}+21\right)\right)$$

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.

$$x^{2}-42x+45=0$$

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.

$$x=\frac{-\left(-42\right)±\sqrt{\left(-42\right)^{2}-4\times 45}}{2}$$

-42 kvadratini chiqarish.

$$x=\frac{-\left(-42\right)±\sqrt{1764-4\times 45}}{2}$$

-4 ni 45 marotabaga ko'paytirish.

$$x=\frac{-\left(-42\right)±\sqrt{1764-180}}{2}$$

1764 ni -180 ga qo'shish.

$$x=\frac{-\left(-42\right)±\sqrt{1584}}{2}$$

1584 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-42\right)±12\sqrt{11}}{2}$$

-42 ning teskarisi 42 ga teng.

$$x=\frac{42±12\sqrt{11}}{2}$$

x=\frac{42±12\sqrt{11}}{2} tenglamasini yeching, bunda ± musbat. 42 ni 12\sqrt{11} ga qo'shish.

$$x=\frac{12\sqrt{11}+42}{2}$$

42+12\sqrt{11} ni 2 ga bo'lish.

$$x=6\sqrt{11}+21$$

x=\frac{42±12\sqrt{11}}{2} tenglamasini yeching, bunda ± manfiy. 42 dan 12\sqrt{11} ni ayirish.

$$x=\frac{42-12\sqrt{11}}{2}$$

42-12\sqrt{11} ni 2 ga bo'lish.

$$x=21-6\sqrt{11}$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun 21+6\sqrt{11} ga va x_{2} uchun 21-6\sqrt{11} ga bo‘ling.

$$x^{2}-42x+45=\left(x-\left(6\sqrt{11}+21\right)\right)\left(x-\left(21-6\sqrt{11}\right)\right)$$

Baholash

$$x^{2}-42x+45$$