3x^2+15x-12=0 eching

$$3 x ^ { 2 } + 15 x - 12 = 0$$

x uchun yechish

$$x=\frac{\sqrt{41}-5}{2}\approx 0,701562119$$

$$x=\frac{-\sqrt{41}-5}{2}\approx -5,701562119$$

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.

$$3x^{2}+15x-12=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 3 ni a, 15 ni b va -12 ni c bilan almashtiring.

$$x=\frac{-15±\sqrt{15^{2}-4\times 3\left(-12\right)}}{2\times 3}$$

15 kvadratini chiqarish.

$$x=\frac{-15±\sqrt{225-4\times 3\left(-12\right)}}{2\times 3}$$

-4 ni 3 marotabaga ko'paytirish.

$$x=\frac{-15±\sqrt{225-12\left(-12\right)}}{2\times 3}$$

-12 ni -12 marotabaga ko'paytirish.

$$x=\frac{-15±\sqrt{225+144}}{2\times 3}$$

225 ni 144 ga qo'shish.

$$x=\frac{-15±\sqrt{369}}{2\times 3}$$

369 ning kvadrat ildizini chiqarish.

$$x=\frac{-15±3\sqrt{41}}{2\times 3}$$

2 ni 3 marotabaga ko'paytirish.

$$x=\frac{-15±3\sqrt{41}}{6}$$

x=\frac{-15±3\sqrt{41}}{6} tenglamasini yeching, bunda ± musbat. -15 ni 3\sqrt{41} ga qo'shish.

$$x=\frac{3\sqrt{41}-15}{6}$$

-15+3\sqrt{41} ni 6 ga bo'lish.

$$x=\frac{\sqrt{41}-5}{2}$$

x=\frac{-15±3\sqrt{41}}{6} tenglamasini yeching, bunda ± manfiy. -15 dan 3\sqrt{41} ni ayirish.

$$x=\frac{-3\sqrt{41}-15}{6}$$

-15-3\sqrt{41} ni 6 ga bo'lish.

$$x=\frac{-\sqrt{41}-5}{2}$$

Tenglama yechildi.

$$x=\frac{\sqrt{41}-5}{2} x=\frac{-\sqrt{41}-5}{2}$$

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.

$$3x^{2}+15x-12=0$$

12 ni tenglamaning ikkala tarafiga qo'shish.

$$3x^{2}+15x-12-\left(-12\right)=-\left(-12\right)$$

O‘zidan -12 ayirilsa 0 qoladi.

$$3x^{2}+15x=-\left(-12\right)$$

0 dan -12 ni ayirish.

$$3x^{2}+15x=12$$

Ikki tarafini 3 ga bo‘ling.

$$\frac{3x^{2}+15x}{3}=\frac{12}{3}$$

3 ga bo'lish 3 ga ko'paytirishni bekor qiladi.

$$x^{2}+\frac{15}{3}x=\frac{12}{3}$$

15 ni 3 ga bo'lish.

$$x^{2}+5x=\frac{12}{3}$$

12 ni 3 ga bo'lish.

$$x^{2}+5x=4$$

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.

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

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

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

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

$$x^{2}+5x+\frac{25}{4}=\frac{41}{4}$$

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.

$$\left(x+\frac{5}{2}\right)^{2}=\frac{41}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{5}{2}\right)^{2}}=\sqrt{\frac{41}{4}}$$

Qisqartirish.

$$x+\frac{5}{2}=\frac{\sqrt{41}}{2} x+\frac{5}{2}=-\frac{\sqrt{41}}{2}$$

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

$$x=\frac{\sqrt{41}-5}{2} x=\frac{-\sqrt{41}-5}{2}$$