2x^2-9x+5=0 eching

$$2 x ^ { 2 } - 9 x + 5 = 0$$

x uchun yechish

$$x = \frac{\sqrt{41} + 9}{4} \approx 3,850781059$$

$$x=\frac{9-\sqrt{41}}{4}\approx 0,649218941$$

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.

$$2x^{2}-9x+5=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 2 ni a, -9 ni b va 5 ni c bilan almashtiring.

$$x=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 2\times 5}}{2\times 2}$$

-9 kvadratini chiqarish.

$$x=\frac{-\left(-9\right)±\sqrt{81-4\times 2\times 5}}{2\times 2}$$

-4 ni 2 marotabaga ko'paytirish.

$$x=\frac{-\left(-9\right)±\sqrt{81-8\times 5}}{2\times 2}$$

-8 ni 5 marotabaga ko'paytirish.

$$x=\frac{-\left(-9\right)±\sqrt{81-40}}{2\times 2}$$

81 ni -40 ga qo'shish.

$$x=\frac{-\left(-9\right)±\sqrt{41}}{2\times 2}$$

-9 ning teskarisi 9 ga teng.

$$x=\frac{9±\sqrt{41}}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$x=\frac{9±\sqrt{41}}{4}$$

x=\frac{9±\sqrt{41}}{4} tenglamasini yeching, bunda ± musbat. 9 ni \sqrt{41} ga qo'shish.

$$x=\frac{\sqrt{41}+9}{4}$$

x=\frac{9±\sqrt{41}}{4} tenglamasini yeching, bunda ± manfiy. 9 dan \sqrt{41} ni ayirish.

$$x=\frac{9-\sqrt{41}}{4}$$

Tenglama yechildi.

$$x=\frac{\sqrt{41}+9}{4} x=\frac{9-\sqrt{41}}{4}$$

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.

$$2x^{2}-9x+5=0$$

Tenglamaning ikkala tarafidan 5 ni ayirish.

$$2x^{2}-9x+5-5=-5$$

O‘zidan 5 ayirilsa 0 qoladi.

$$2x^{2}-9x=-5$$

Ikki tarafini 2 ga bo‘ling.

$$\frac{2x^{2}-9x}{2}=-\frac{5}{2}$$

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

$$x^{2}-\frac{9}{2}x=-\frac{5}{2}$$

-\frac{9}{2} ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{9}{4} olish uchun. Keyin, -\frac{9}{4} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}-\frac{9}{2}x+\left(-\frac{9}{4}\right)^{2}=-\frac{5}{2}+\left(-\frac{9}{4}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{9}{4} kvadratini chiqarish.

$$x^{2}-\frac{9}{2}x+\frac{81}{16}=-\frac{5}{2}+\frac{81}{16}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali -\frac{5}{2} ni \frac{81}{16} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}-\frac{9}{2}x+\frac{81}{16}=\frac{41}{16}$$

x^{2}-\frac{9}{2}x+\frac{81}{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.

$$\left(x-\frac{9}{4}\right)^{2}=\frac{41}{16}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{9}{4}\right)^{2}}=\sqrt{\frac{41}{16}}$$

Qisqartirish.

$$x-\frac{9}{4}=\frac{\sqrt{41}}{4} x-\frac{9}{4}=-\frac{\sqrt{41}}{4}$$

\frac{9}{4} ni tenglamaning ikkala tarafiga qo'shish.

$$x=\frac{\sqrt{41}+9}{4} x=\frac{9-\sqrt{41}}{4}$$