x^2-7x+2 eching

$${ x }^{ 2 } -7x+2$$

Omil

$$\left(x-\frac{7-\sqrt{41}}{2}\right)\left(x-\frac{\sqrt{41}+7}{2}\right)$$

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.

$$x^{2}-7x+2=0$$

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.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 2}}{2}$$

-7 kvadratini chiqarish.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\times 2}}{2}$$

-4 ni 2 marotabaga ko'paytirish.

$$x=\frac{-\left(-7\right)±\sqrt{49-8}}{2}$$

49 ni -8 ga qo'shish.

$$x=\frac{-\left(-7\right)±\sqrt{41}}{2}$$

-7 ning teskarisi 7 ga teng.

$$x=\frac{7±\sqrt{41}}{2}$$

x=\frac{7±\sqrt{41}}{2} tenglamasini yeching, bunda ± musbat. 7 ni \sqrt{41} ga qo'shish.

$$x=\frac{\sqrt{41}+7}{2}$$

x=\frac{7±\sqrt{41}}{2} tenglamasini yeching, bunda ± manfiy. 7 dan \sqrt{41} ni ayirish.

$$x=\frac{7-\sqrt{41}}{2}$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun \frac{7+\sqrt{41}}{2} ga va x_{2} uchun \frac{7-\sqrt{41}}{2} ga bo‘ling.

$$x^{2}-7x+2=\left(x-\frac{\sqrt{41}+7}{2}\right)\left(x-\frac{7-\sqrt{41}}{2}\right)$$

Baholash

$$x^{2}-7x+2$$