2x^2+12x-9=0 eching

$$2 { x }^{ 2 } +12x-9=0$$

x uchun yechish

$$x=\frac{3\sqrt{6}}{2}-3\approx 0,674234614$$

$$x=-\frac{3\sqrt{6}}{2}-3\approx -6,674234614$$

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.

$$2x^{2}+12x-9=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 2 ni a, 12 ni b va -9 ni c bilan almashtiring.

$$x=\frac{-12±\sqrt{12^{2}-4\times 2\left(-9\right)}}{2\times 2}$$

12 kvadratini chiqarish.

$$x=\frac{-12±\sqrt{144-4\times 2\left(-9\right)}}{2\times 2}$$

-4 ni 2 marotabaga ko'paytirish.

$$x=\frac{-12±\sqrt{144-8\left(-9\right)}}{2\times 2}$$

-8 ni -9 marotabaga ko'paytirish.

$$x=\frac{-12±\sqrt{144+72}}{2\times 2}$$

144 ni 72 ga qo'shish.

$$x=\frac{-12±\sqrt{216}}{2\times 2}$$

216 ning kvadrat ildizini chiqarish.

$$x=\frac{-12±6\sqrt{6}}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$x=\frac{-12±6\sqrt{6}}{4}$$

x=\frac{-12±6\sqrt{6}}{4} tenglamasini yeching, bunda ± musbat. -12 ni 6\sqrt{6} ga qo'shish.

$$x=\frac{6\sqrt{6}-12}{4}$$

-12+6\sqrt{6} ni 4 ga bo'lish.

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

x=\frac{-12±6\sqrt{6}}{4} tenglamasini yeching, bunda ± manfiy. -12 dan 6\sqrt{6} ni ayirish.

$$x=\frac{-6\sqrt{6}-12}{4}$$

-12-6\sqrt{6} ni 4 ga bo'lish.

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

Tenglama yechildi.

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

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.

$$2x^{2}+12x-9=0$$

9 ni tenglamaning ikkala tarafiga qo'shish.

$$2x^{2}+12x-9-\left(-9\right)=-\left(-9\right)$$

O‘zidan -9 ayirilsa 0 qoladi.

$$2x^{2}+12x=-\left(-9\right)$$

0 dan -9 ni ayirish.

$$2x^{2}+12x=9$$

Ikki tarafini 2 ga bo‘ling.

$$\frac{2x^{2}+12x}{2}=\frac{9}{2}$$

2 ga bo'lish 2 ga ko'paytirishni bekor qiladi.

$$x^{2}+\frac{12}{2}x=\frac{9}{2}$$

12 ni 2 ga bo'lish.

$$x^{2}+6x=\frac{9}{2}$$

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.

$$x^{2}+6x+3^{2}=\frac{9}{2}+3^{2}$$

3 kvadratini chiqarish.

$$x^{2}+6x+9=\frac{9}{2}+9$$

\frac{9}{2} ni 9 ga qo'shish.

$$x^{2}+6x+9=\frac{27}{2}$$

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.

$$\left(x+3\right)^{2}=\frac{27}{2}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+3\right)^{2}}=\sqrt{\frac{27}{2}}$$

Qisqartirish.

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

Tenglamaning ikkala tarafidan 3 ni ayirish.

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