x^2-38x+9 eching

$$x ^ { 2 } - 38 x + 9$$

Omil

$$\left(x-\left(19-4\sqrt{22}\right)\right)\left(x-\left(4\sqrt{22}+19\right)\right)$$

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.

$$x^{2}-38x+9=0$$

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.

$$x=\frac{-\left(-38\right)±\sqrt{\left(-38\right)^{2}-4\times 9}}{2}$$

-38 kvadratini chiqarish.

$$x=\frac{-\left(-38\right)±\sqrt{1444-4\times 9}}{2}$$

-4 ni 9 marotabaga ko'paytirish.

$$x=\frac{-\left(-38\right)±\sqrt{1444-36}}{2}$$

1444 ni -36 ga qo'shish.

$$x=\frac{-\left(-38\right)±\sqrt{1408}}{2}$$

1408 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-38\right)±8\sqrt{22}}{2}$$

-38 ning teskarisi 38 ga teng.

$$x=\frac{38±8\sqrt{22}}{2}$$

x=\frac{38±8\sqrt{22}}{2} tenglamasini yeching, bunda ± musbat. 38 ni 8\sqrt{22} ga qo'shish.

$$x=\frac{8\sqrt{22}+38}{2}$$

38+8\sqrt{22} ni 2 ga bo'lish.

$$x=4\sqrt{22}+19$$

x=\frac{38±8\sqrt{22}}{2} tenglamasini yeching, bunda ± manfiy. 38 dan 8\sqrt{22} ni ayirish.

$$x=\frac{38-8\sqrt{22}}{2}$$

38-8\sqrt{22} ni 2 ga bo'lish.

$$x=19-4\sqrt{22}$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun 19+4\sqrt{22} ga va x_{2} uchun 19-4\sqrt{22} ga bo‘ling.

$$x^{2}-38x+9=\left(x-\left(4\sqrt{22}+19\right)\right)\left(x-\left(19-4\sqrt{22}\right)\right)$$

Baholash

$$x^{2}-38x+9$$