x^2+6x-5=0 eching

$${ x }^{ 2 } +6x-5=0$$

x uchun yechish (complex solution)

$$x=\sqrt{14}-3\approx 0,741657387$$

$$x=-\left(\sqrt{14}+3\right)\approx -6,741657387$$

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.

$$x^{2}+6x-5=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 1 ni a, 6 ni b va -5 ni c bilan almashtiring.

$$x=\frac{-6±\sqrt{6^{2}-4\left(-5\right)}}{2}$$

6 kvadratini chiqarish.

$$x=\frac{-6±\sqrt{36-4\left(-5\right)}}{2}$$

-4 ni -5 marotabaga ko'paytirish.

$$x=\frac{-6±\sqrt{36+20}}{2}$$

36 ni 20 ga qo'shish.

$$x=\frac{-6±\sqrt{56}}{2}$$

56 ning kvadrat ildizini chiqarish.

$$x=\frac{-6±2\sqrt{14}}{2}$$

x=\frac{-6±2\sqrt{14}}{2} tenglamasini yeching, bunda ± musbat. -6 ni 2\sqrt{14} ga qo'shish.

$$x=\frac{2\sqrt{14}-6}{2}$$

-6+2\sqrt{14} ni 2 ga bo'lish.

$$x=\sqrt{14}-3$$

x=\frac{-6±2\sqrt{14}}{2} tenglamasini yeching, bunda ± manfiy. -6 dan 2\sqrt{14} ni ayirish.

$$x=\frac{-2\sqrt{14}-6}{2}$$

-6-2\sqrt{14} ni 2 ga bo'lish.

$$x=-\sqrt{14}-3$$

Tenglama yechildi.

$$x=\sqrt{14}-3 x=-\sqrt{14}-3$$

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.

$$x^{2}+6x-5=0$$

5 ni tenglamaning ikkala tarafiga qo'shish.

$$x^{2}+6x-5-\left(-5\right)=-\left(-5\right)$$

O‘zidan -5 ayirilsa 0 qoladi.

$$x^{2}+6x=-\left(-5\right)$$

0 dan -5 ni ayirish.

$$x^{2}+6x=5$$

6 ni bo‘lish, x shartining koeffitsienti, 2 ga 3 olish uchun. Keyin, 3 ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+6x+3^{2}=5+3^{2}$$

3 kvadratini chiqarish.

$$x^{2}+6x+9=5+9$$

5 ni 9 ga qo'shish.

$$x^{2}+6x+9=14$$

x^{2}+6x+9 omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+3\right)^{2}=14$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+3\right)^{2}}=\sqrt{14}$$

Qisqartirish.

$$x+3=\sqrt{14} x+3=-\sqrt{14}$$

Tenglamaning ikkala tarafidan 3 ni ayirish.

$$x=\sqrt{14}-3 x=-\sqrt{14}-3$$

x uchun yechish

$$x=\sqrt{14}-3\approx 0,741657387$$

$$x=-\sqrt{14}-3\approx -6,741657387$$

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.

$$x^{2}+6x-5=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 1 ni a, 6 ni b va -5 ni c bilan almashtiring.

$$x=\frac{-6±\sqrt{6^{2}-4\left(-5\right)}}{2}$$

6 kvadratini chiqarish.

$$x=\frac{-6±\sqrt{36-4\left(-5\right)}}{2}$$

-4 ni -5 marotabaga ko'paytirish.

$$x=\frac{-6±\sqrt{36+20}}{2}$$

36 ni 20 ga qo'shish.

$$x=\frac{-6±\sqrt{56}}{2}$$

56 ning kvadrat ildizini chiqarish.

$$x=\frac{-6±2\sqrt{14}}{2}$$

x=\frac{-6±2\sqrt{14}}{2} tenglamasini yeching, bunda ± musbat. -6 ni 2\sqrt{14} ga qo'shish.

$$x=\frac{2\sqrt{14}-6}{2}$$

-6+2\sqrt{14} ni 2 ga bo'lish.

$$x=\sqrt{14}-3$$

x=\frac{-6±2\sqrt{14}}{2} tenglamasini yeching, bunda ± manfiy. -6 dan 2\sqrt{14} ni ayirish.

$$x=\frac{-2\sqrt{14}-6}{2}$$

-6-2\sqrt{14} ni 2 ga bo'lish.

$$x=-\sqrt{14}-3$$

Tenglama yechildi.

$$x=\sqrt{14}-3 x=-\sqrt{14}-3$$

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.

$$x^{2}+6x-5=0$$

5 ni tenglamaning ikkala tarafiga qo'shish.

$$x^{2}+6x-5-\left(-5\right)=-\left(-5\right)$$

O‘zidan -5 ayirilsa 0 qoladi.

$$x^{2}+6x=-\left(-5\right)$$

0 dan -5 ni ayirish.

$$x^{2}+6x=5$$

6 ni bo‘lish, x shartining koeffitsienti, 2 ga 3 olish uchun. Keyin, 3 ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+6x+3^{2}=5+3^{2}$$

3 kvadratini chiqarish.

$$x^{2}+6x+9=5+9$$

5 ni 9 ga qo'shish.

$$x^{2}+6x+9=14$$

x^{2}+6x+9 omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+3\right)^{2}=14$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+3\right)^{2}}=\sqrt{14}$$

Qisqartirish.

$$x+3=\sqrt{14} x+3=-\sqrt{14}$$

Tenglamaning ikkala tarafidan 3 ni ayirish.

$$x=\sqrt{14}-3 x=-\sqrt{14}-3$$