x^2+9x+5=0 eching

$$x ^ { 2 } + 9 x + 5 = 0$$

x uchun yechish

$$x=\frac{\sqrt{61}-9}{2}\approx -0,594875162$$

$$x=\frac{-\sqrt{61}-9}{2}\approx -8,405124838$$

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.

$$x^{2}+9x+5=0$$

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 5 ni c bilan almashtiring.

$$x=\frac{-9±\sqrt{9^{2}-4\times 5}}{2}$$

9 kvadratini chiqarish.

$$x=\frac{-9±\sqrt{81-4\times 5}}{2}$$

-4 ni 5 marotabaga ko'paytirish.

$$x=\frac{-9±\sqrt{81-20}}{2}$$

81 ni -20 ga qo'shish.

$$x=\frac{-9±\sqrt{61}}{2}$$

x=\frac{-9±\sqrt{61}}{2} tenglamasini yeching, bunda ± musbat. -9 ni \sqrt{61} ga qo'shish.

$$x=\frac{\sqrt{61}-9}{2}$$

x=\frac{-9±\sqrt{61}}{2} tenglamasini yeching, bunda ± manfiy. -9 dan \sqrt{61} ni ayirish.

$$x=\frac{-\sqrt{61}-9}{2}$$

Tenglama yechildi.

$$x=\frac{\sqrt{61}-9}{2} x=\frac{-\sqrt{61}-9}{2}$$

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.

$$x^{2}+9x+5=0$$

Tenglamaning ikkala tarafidan 5 ni ayirish.

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

O‘zidan 5 ayirilsa 0 qoladi.

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

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.

$$x^{2}+9x+\left(\frac{9}{2}\right)^{2}=-5+\left(\frac{9}{2}\right)^{2}$$

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

$$x^{2}+9x+\frac{81}{4}=-5+\frac{81}{4}$$

-5 ni \frac{81}{4} ga qo'shish.

$$x^{2}+9x+\frac{81}{4}=\frac{61}{4}$$

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.

$$\left(x+\frac{9}{2}\right)^{2}=\frac{61}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{9}{2}\right)^{2}}=\sqrt{\frac{61}{4}}$$

Qisqartirish.

$$x+\frac{9}{2}=\frac{\sqrt{61}}{2} x+\frac{9}{2}=-\frac{\sqrt{61}}{2}$$

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

$$x=\frac{\sqrt{61}-9}{2} x=\frac{-\sqrt{61}-9}{2}$$