4x^2-52x-68 eching

$$4 x ^ { 2 } - 52 x - 68$$

Omil

$$4\left(x-\frac{13-\sqrt{237}}{2}\right)\left(x-\frac{\sqrt{237}+13}{2}\right)$$

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.

$$4x^{2}-52x-68=0$$

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.

$$x=\frac{-\left(-52\right)±\sqrt{\left(-52\right)^{2}-4\times 4\left(-68\right)}}{2\times 4}$$

-52 kvadratini chiqarish.

$$x=\frac{-\left(-52\right)±\sqrt{2704-4\times 4\left(-68\right)}}{2\times 4}$$

-4 ni 4 marotabaga ko'paytirish.

$$x=\frac{-\left(-52\right)±\sqrt{2704-16\left(-68\right)}}{2\times 4}$$

-16 ni -68 marotabaga ko'paytirish.

$$x=\frac{-\left(-52\right)±\sqrt{2704+1088}}{2\times 4}$$

2704 ni 1088 ga qo'shish.

$$x=\frac{-\left(-52\right)±\sqrt{3792}}{2\times 4}$$

3792 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-52\right)±4\sqrt{237}}{2\times 4}$$

-52 ning teskarisi 52 ga teng.

$$x=\frac{52±4\sqrt{237}}{2\times 4}$$

2 ni 4 marotabaga ko'paytirish.

$$x=\frac{52±4\sqrt{237}}{8}$$

x=\frac{52±4\sqrt{237}}{8} tenglamasini yeching, bunda ± musbat. 52 ni 4\sqrt{237} ga qo'shish.

$$x=\frac{4\sqrt{237}+52}{8}$$

52+4\sqrt{237} ni 8 ga bo'lish.

$$x=\frac{\sqrt{237}+13}{2}$$

x=\frac{52±4\sqrt{237}}{8} tenglamasini yeching, bunda ± manfiy. 52 dan 4\sqrt{237} ni ayirish.

$$x=\frac{52-4\sqrt{237}}{8}$$

52-4\sqrt{237} ni 8 ga bo'lish.

$$x=\frac{13-\sqrt{237}}{2}$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun \frac{13+\sqrt{237}}{2} ga va x_{2} uchun \frac{13-\sqrt{237}}{2} ga bo‘ling.

$$4x^{2}-52x-68=4\left(x-\frac{\sqrt{237}+13}{2}\right)\left(x-\frac{13-\sqrt{237}}{2}\right)$$

Baholash

$$4\left(x^{2}-13x-17\right)$$