x^2+8+6x=89 eching

$${ x }^{ 2 } +8+6x=89$$

x uchun yechish

$$x=3\sqrt{10}-3\approx 6,486832981$$

$$x=-3\sqrt{10}-3\approx -12,486832981$$

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.

$$x^{2}+6x+8=89$$

Tenglamaning ikkala tarafidan 89 ni ayirish.

$$x^{2}+6x+8-89=89-89$$

O‘zidan 89 ayirilsa 0 qoladi.

$$x^{2}+6x+8-89=0$$

8 dan 89 ni ayirish.

$$x^{2}+6x-81=0$$

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 -81 ni c bilan almashtiring.

$$x=\frac{-6±\sqrt{6^{2}-4\left(-81\right)}}{2}$$

6 kvadratini chiqarish.

$$x=\frac{-6±\sqrt{36-4\left(-81\right)}}{2}$$

-4 ni -81 marotabaga ko'paytirish.

$$x=\frac{-6±\sqrt{36+324}}{2}$$

36 ni 324 ga qo'shish.

$$x=\frac{-6±\sqrt{360}}{2}$$

360 ning kvadrat ildizini chiqarish.

$$x=\frac{-6±6\sqrt{10}}{2}$$

x=\frac{-6±6\sqrt{10}}{2} tenglamasini yeching, bunda ± musbat. -6 ni 6\sqrt{10} ga qo'shish.

$$x=\frac{6\sqrt{10}-6}{2}$$

-6+6\sqrt{10} ni 2 ga bo'lish.

$$x=3\sqrt{10}-3$$

x=\frac{-6±6\sqrt{10}}{2} tenglamasini yeching, bunda ± manfiy. -6 dan 6\sqrt{10} ni ayirish.

$$x=\frac{-6\sqrt{10}-6}{2}$$

-6-6\sqrt{10} ni 2 ga bo'lish.

$$x=-3\sqrt{10}-3$$

Tenglama yechildi.

$$x=3\sqrt{10}-3 x=-3\sqrt{10}-3$$

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.

$$x^{2}+6x+8=89$$

Tenglamaning ikkala tarafidan 8 ni ayirish.

$$x^{2}+6x+8-8=89-8$$

O‘zidan 8 ayirilsa 0 qoladi.

$$x^{2}+6x=89-8$$

89 dan 8 ni ayirish.

$$x^{2}+6x=81$$

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.

$$x^{2}+6x+3^{2}=81+3^{2}$$

3 kvadratini chiqarish.

$$x^{2}+6x+9=81+9$$

81 ni 9 ga qo'shish.

$$x^{2}+6x+9=90$$

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.

$$\left(x+3\right)^{2}=90$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+3\right)^{2}}=\sqrt{90}$$

Qisqartirish.

$$x+3=3\sqrt{10} x+3=-3\sqrt{10}$$

Tenglamaning ikkala tarafidan 3 ni ayirish.

$$x=3\sqrt{10}-3 x=-3\sqrt{10}-3$$