x^2+9x+9=0 eching

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

x uchun yechish

$$x=\frac{3\sqrt{5}-9}{2}\approx -1,145898034$$

$$x=\frac{-3\sqrt{5}-9}{2}\approx -7,854101966$$

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.

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

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 1 ni a, 9 ni b va 9 ni c bilan almashtiring.

$$x=\frac{-9±\sqrt{9^{2}-4\times 9}}{2}$$

9 kvadratini chiqarish.

$$x=\frac{-9±\sqrt{81-4\times 9}}{2}$$

-4 ni 9 marotabaga ko'paytirish.

$$x=\frac{-9±\sqrt{81-36}}{2}$$

81 ni -36 ga qo'shish.

$$x=\frac{-9±\sqrt{45}}{2}$$

45 ning kvadrat ildizini chiqarish.

$$x=\frac{-9±3\sqrt{5}}{2}$$

x=\frac{-9±3\sqrt{5}}{2} tenglamasini yeching, bunda ± musbat. -9 ni 3\sqrt{5} ga qo'shish.

$$x=\frac{3\sqrt{5}-9}{2}$$

x=\frac{-9±3\sqrt{5}}{2} tenglamasini yeching, bunda ± manfiy. -9 dan 3\sqrt{5} ni ayirish.

$$x=\frac{-3\sqrt{5}-9}{2}$$

Tenglama yechildi.

$$x=\frac{3\sqrt{5}-9}{2} x=\frac{-3\sqrt{5}-9}{2}$$

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.

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

Tenglamaning ikkala tarafidan 9 ni ayirish.

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

O‘zidan 9 ayirilsa 0 qoladi.

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

9 ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{9}{2} olish uchun. Keyin, \frac{9}{2} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

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

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{9}{2} kvadratini chiqarish.

$$x^{2}+9x+\frac{81}{4}=-9+\frac{81}{4}$$

-9 ni \frac{81}{4} ga qo'shish.

$$x^{2}+9x+\frac{81}{4}=\frac{45}{4}$$

x^{2}+9x+\frac{81}{4} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+\frac{9}{2}\right)^{2}=\frac{45}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{9}{2}\right)^{2}}=\sqrt{\frac{45}{4}}$$

Qisqartirish.

$$x+\frac{9}{2}=\frac{3\sqrt{5}}{2} x+\frac{9}{2}=-\frac{3\sqrt{5}}{2}$$

Tenglamaning ikkala tarafidan \frac{9}{2} ni ayirish.

$$x=\frac{3\sqrt{5}-9}{2} x=\frac{-3\sqrt{5}-9}{2}$$