x^2+2x-5/3=0 eching

$$x ^ { 2 } + 2 x - \frac { 5 } { 3 } = 0$$

x uchun yechish

$$x=\frac{2\sqrt{6}}{3}-1\approx 0,632993162$$

$$x=-\frac{2\sqrt{6}}{3}-1\approx -2,632993162$$

Kvadrat tenglamalaridan foydalanish qoidalari

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.

$$x^{2}+2x-\frac{5}{3}=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 1 ni a, 2 ni b va -\frac{5}{3} ni c bilan almashtiring.

$$x=\frac{-2±\sqrt{2^{2}-4\left(-\frac{5}{3}\right)}}{2}$$

2 kvadratini chiqarish.

$$x=\frac{-2±\sqrt{4-4\left(-\frac{5}{3}\right)}}{2}$$

-4 ni -\frac{5}{3} marotabaga ko'paytirish.

$$x=\frac{-2±\sqrt{4+\frac{20}{3}}}{2}$$

4 ni \frac{20}{3} ga qo'shish.

$$x=\frac{-2±\sqrt{\frac{32}{3}}}{2}$$

\frac{32}{3} ning kvadrat ildizini chiqarish.

$$x=\frac{-2±\frac{4\sqrt{6}}{3}}{2}$$

x=\frac{-2±\frac{4\sqrt{6}}{3}}{2} tenglamasini yeching, bunda ± musbat. -2 ni \frac{4\sqrt{6}}{3} ga qo'shish.

$$x=\frac{\frac{4\sqrt{6}}{3}-2}{2}$$

-2+\frac{4\sqrt{6}}{3} ni 2 ga bo'lish.

$$x=\frac{2\sqrt{6}}{3}-1$$

x=\frac{-2±\frac{4\sqrt{6}}{3}}{2} tenglamasini yeching, bunda ± manfiy. -2 dan \frac{4\sqrt{6}}{3} ni ayirish.

$$x=\frac{-\frac{4\sqrt{6}}{3}-2}{2}$$

-2-\frac{4\sqrt{6}}{3} ni 2 ga bo'lish.

$$x=-\frac{2\sqrt{6}}{3}-1$$

Tenglama yechildi.

$$x=\frac{2\sqrt{6}}{3}-1 x=-\frac{2\sqrt{6}}{3}-1$$

Kvadratni yakunlash bo'yicha qadamlar

Bu kabi kvadrat tenglamalarni kvadratni yakunlab yechish mumkin. Kvadratni yechish uchun tenglama avval ushbu shaklda bo'lishi shart: x^{2}+bx=c.

$$x^{2}+2x-\frac{5}{3}=0$$

\frac{5}{3} ni tenglamaning ikkala tarafiga qo'shish.

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

O‘zidan -\frac{5}{3} ayirilsa 0 qoladi.

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

0 dan -\frac{5}{3} ni ayirish.

$$x^{2}+2x=\frac{5}{3}$$

2 ni bo‘lish, x shartining koeffitsienti, 2 ga 1 olish uchun. Keyin, 1 ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+2x+1^{2}=\frac{5}{3}+1^{2}$$

1 kvadratini chiqarish.

$$x^{2}+2x+1=\frac{5}{3}+1$$

\frac{5}{3} ni 1 ga qo'shish.

$$x^{2}+2x+1=\frac{8}{3}$$

x^{2}+2x+1 omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+1\right)^{2}=\frac{8}{3}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{\frac{8}{3}}$$

Qisqartirish.

$$x+1=\frac{2\sqrt{6}}{3} x+1=-\frac{2\sqrt{6}}{3}$$

Tenglamaning ikkala tarafidan 1 ni ayirish.

$$x=\frac{2\sqrt{6}}{3}-1 x=-\frac{2\sqrt{6}}{3}-1$$