x^2+5x+7=0 eching

$${ x }^{ 2 } +5x+7 = 0$$

x uchun yechish (complex solution)

$$x=\frac{-5+\sqrt{3}i}{2}\approx -2,5+0,866025404i$$

$$x=\frac{-\sqrt{3}i-5}{2}\approx -2,5-0,866025404i$$

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.

$$x^{2}+5x+7=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 1 ni a, 5 ni b va 7 ni c bilan almashtiring.

$$x=\frac{-5±\sqrt{5^{2}-4\times 7}}{2}$$

5 kvadratini chiqarish.

$$x=\frac{-5±\sqrt{25-4\times 7}}{2}$$

-4 ni 7 marotabaga ko'paytirish.

$$x=\frac{-5±\sqrt{25-28}}{2}$$

25 ni -28 ga qo'shish.

$$x=\frac{-5±\sqrt{-3}}{2}$$

-3 ning kvadrat ildizini chiqarish.

$$x=\frac{-5±\sqrt{3}i}{2}$$

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

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

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

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

Tenglama yechildi.

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

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.

$$x^{2}+5x+7=0$$

Tenglamaning ikkala tarafidan 7 ni ayirish.

$$x^{2}+5x+7-7=-7$$

O‘zidan 7 ayirilsa 0 qoladi.

$$x^{2}+5x=-7$$

5 ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{5}{2} olish uchun. Keyin, \frac{5}{2} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+5x+\left(\frac{5}{2}\right)^{2}=-7+\left(\frac{5}{2}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{5}{2} kvadratini chiqarish.

$$x^{2}+5x+\frac{25}{4}=-7+\frac{25}{4}$$

-7 ni \frac{25}{4} ga qo'shish.

$$x^{2}+5x+\frac{25}{4}=-\frac{3}{4}$$

x^{2}+5x+\frac{25}{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.

$$\left(x+\frac{5}{2}\right)^{2}=-\frac{3}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{5}{2}\right)^{2}}=\sqrt{-\frac{3}{4}}$$

Qisqartirish.

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

Tenglamaning ikkala tarafidan \frac{5}{2} ni ayirish.

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