2x^2+5x+3=20 eching

$$2 { x }^{ 2 } +5x+3=20$$

x uchun yechish

$$x = \frac{\sqrt{161} - 5}{4} \approx 1,922144385$$

$$x=\frac{-\sqrt{161}-5}{4}\approx -4,422144385$$

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.

$$2x^{2}+5x+3=20$$

Tenglamaning ikkala tarafidan 20 ni ayirish.

$$2x^{2}+5x+3-20=20-20$$

O‘zidan 20 ayirilsa 0 qoladi.

$$2x^{2}+5x+3-20=0$$

3 dan 20 ni ayirish.

$$2x^{2}+5x-17=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 2 ni a, 5 ni b va -17 ni c bilan almashtiring.

$$x=\frac{-5±\sqrt{5^{2}-4\times 2\left(-17\right)}}{2\times 2}$$

5 kvadratini chiqarish.

$$x=\frac{-5±\sqrt{25-4\times 2\left(-17\right)}}{2\times 2}$$

-4 ni 2 marotabaga ko'paytirish.

$$x=\frac{-5±\sqrt{25-8\left(-17\right)}}{2\times 2}$$

-8 ni -17 marotabaga ko'paytirish.

$$x=\frac{-5±\sqrt{25+136}}{2\times 2}$$

25 ni 136 ga qo'shish.

$$x=\frac{-5±\sqrt{161}}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$x=\frac{-5±\sqrt{161}}{4}$$

x=\frac{-5±\sqrt{161}}{4} tenglamasini yeching, bunda ± musbat. -5 ni \sqrt{161} ga qo'shish.

$$x=\frac{\sqrt{161}-5}{4}$$

x=\frac{-5±\sqrt{161}}{4} tenglamasini yeching, bunda ± manfiy. -5 dan \sqrt{161} ni ayirish.

$$x=\frac{-\sqrt{161}-5}{4}$$

Tenglama yechildi.

$$x=\frac{\sqrt{161}-5}{4} x=\frac{-\sqrt{161}-5}{4}$$

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.

$$2x^{2}+5x+3=20$$

Tenglamaning ikkala tarafidan 3 ni ayirish.

$$2x^{2}+5x+3-3=20-3$$

O‘zidan 3 ayirilsa 0 qoladi.

$$2x^{2}+5x=20-3$$

20 dan 3 ni ayirish.

$$2x^{2}+5x=17$$

Ikki tarafini 2 ga bo‘ling.

$$\frac{2x^{2}+5x}{2}=\frac{17}{2}$$

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

$$x^{2}+\frac{5}{2}x=\frac{17}{2}$$

\frac{5}{2} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{5}{4} olish uchun. Keyin, \frac{5}{4} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+\frac{5}{2}x+\left(\frac{5}{4}\right)^{2}=\frac{17}{2}+\left(\frac{5}{4}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{5}{4} kvadratini chiqarish.

$$x^{2}+\frac{5}{2}x+\frac{25}{16}=\frac{17}{2}+\frac{25}{16}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{17}{2} ni \frac{25}{16} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}+\frac{5}{2}x+\frac{25}{16}=\frac{161}{16}$$

x^{2}+\frac{5}{2}x+\frac{25}{16} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+\frac{5}{4}\right)^{2}=\frac{161}{16}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{5}{4}\right)^{2}}=\sqrt{\frac{161}{16}}$$

Qisqartirish.

$$x+\frac{5}{4}=\frac{\sqrt{161}}{4} x+\frac{5}{4}=-\frac{\sqrt{161}}{4}$$

Tenglamaning ikkala tarafidan \frac{5}{4} ni ayirish.

$$x=\frac{\sqrt{161}-5}{4} x=\frac{-\sqrt{161}-5}{4}$$