-2x^2-5x+5=0 eching

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

x uchun yechish

$$x=\frac{\sqrt{65}-5}{4}\approx 0,765564437$$

$$x=\frac{-\sqrt{65}-5}{4}\approx -3,265564437$$

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.

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

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 5 ni c bilan almashtiring.

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

-5 kvadratini chiqarish.

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

-4 ni -2 marotabaga ko'paytirish.

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

8 ni 5 marotabaga ko'paytirish.

$$x=\frac{-\left(-5\right)±\sqrt{25+40}}{2\left(-2\right)}$$

25 ni 40 ga qo'shish.

$$x=\frac{-\left(-5\right)±\sqrt{65}}{2\left(-2\right)}$$

-5 ning teskarisi 5 ga teng.

$$x=\frac{5±\sqrt{65}}{2\left(-2\right)}$$

2 ni -2 marotabaga ko'paytirish.

$$x=\frac{5±\sqrt{65}}{-4}$$

x=\frac{5±\sqrt{65}}{-4} tenglamasini yeching, bunda ± musbat. 5 ni \sqrt{65} ga qo'shish.

$$x=\frac{\sqrt{65}+5}{-4}$$

5+\sqrt{65} ni -4 ga bo'lish.

$$x=\frac{-\sqrt{65}-5}{4}$$

x=\frac{5±\sqrt{65}}{-4} tenglamasini yeching, bunda ± manfiy. 5 dan \sqrt{65} ni ayirish.

$$x=\frac{5-\sqrt{65}}{-4}$$

5-\sqrt{65} ni -4 ga bo'lish.

$$x=\frac{\sqrt{65}-5}{4}$$

Tenglama yechildi.

$$x=\frac{-\sqrt{65}-5}{4} x=\frac{\sqrt{65}-5}{4}$$

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.

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

Tenglamaning ikkala tarafidan 5 ni ayirish.

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

O‘zidan 5 ayirilsa 0 qoladi.

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

Ikki tarafini -2 ga bo‘ling.

$$\frac{-2x^{2}-5x}{-2}=-\frac{5}{-2}$$

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

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

-5 ni -2 ga bo'lish.

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

-5 ni -2 ga bo'lish.

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

\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.

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

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

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

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{5}{2} ni \frac{25}{16} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}+\frac{5}{2}x+\frac{25}{16}=\frac{65}{16}$$

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.

$$\left(x+\frac{5}{4}\right)^{2}=\frac{65}{16}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{5}{4}\right)^{2}}=\sqrt{\frac{65}{16}}$$

Qisqartirish.

$$x+\frac{5}{4}=\frac{\sqrt{65}}{4} x+\frac{5}{4}=-\frac{\sqrt{65}}{4}$$

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

$$x=\frac{\sqrt{65}-5}{4} x=\frac{-\sqrt{65}-5}{4}$$