2x^2+5x=8 eching

$$2 x ^ { 2 } + 5 x = 8$$

x uchun yechish

$$x = \frac{\sqrt{89} - 5}{4} \approx 1,108495283$$

$$x=\frac{-\sqrt{89}-5}{4}\approx -3,608495283$$

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.

$$2x^{2}+5x=8$$

Tenglamaning ikkala tarafidan 8 ni ayirish.

$$2x^{2}+5x-8=8-8$$

O‘zidan 8 ayirilsa 0 qoladi.

$$2x^{2}+5x-8=0$$

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 -8 ni c bilan almashtiring.

$$x=\frac{-5±\sqrt{5^{2}-4\times 2\left(-8\right)}}{2\times 2}$$

5 kvadratini chiqarish.

$$x=\frac{-5±\sqrt{25-4\times 2\left(-8\right)}}{2\times 2}$$

-4 ni 2 marotabaga ko'paytirish.

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

-8 ni -8 marotabaga ko'paytirish.

$$x=\frac{-5±\sqrt{25+64}}{2\times 2}$$

25 ni 64 ga qo'shish.

$$x=\frac{-5±\sqrt{89}}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$x=\frac{-5±\sqrt{89}}{4}$$

x=\frac{-5±\sqrt{89}}{4} tenglamasini yeching, bunda ± musbat. -5 ni \sqrt{89} ga qo'shish.

$$x=\frac{\sqrt{89}-5}{4}$$

x=\frac{-5±\sqrt{89}}{4} tenglamasini yeching, bunda ± manfiy. -5 dan \sqrt{89} ni ayirish.

$$x=\frac{-\sqrt{89}-5}{4}$$

Tenglama yechildi.

$$x=\frac{\sqrt{89}-5}{4} x=\frac{-\sqrt{89}-5}{4}$$

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.

$$2x^{2}+5x=8$$

Ikki tarafini 2 ga bo‘ling.

$$\frac{2x^{2}+5x}{2}=\frac{8}{2}$$

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

$$x^{2}+\frac{5}{2}x=\frac{8}{2}$$

8 ni 2 ga bo'lish.

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

\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.

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

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

$$x^{2}+\frac{5}{2}x+\frac{25}{16}=4+\frac{25}{16}$$

4 ni \frac{25}{16} ga qo'shish.

$$x^{2}+\frac{5}{2}x+\frac{25}{16}=\frac{89}{16}$$

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.

$$\left(x+\frac{5}{4}\right)^{2}=\frac{89}{16}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{5}{4}\right)^{2}}=\sqrt{\frac{89}{16}}$$

Qisqartirish.

$$x+\frac{5}{4}=\frac{\sqrt{89}}{4} x+\frac{5}{4}=-\frac{\sqrt{89}}{4}$$

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

$$x=\frac{\sqrt{89}-5}{4} x=\frac{-\sqrt{89}-5}{4}$$