w^2-5w=3 eching

$$w ^ { 2 } - 5 w = 3$$

w uchun yechish

$$w = \frac{\sqrt{37} + 5}{2} \approx 5,541381265$$

$$w=\frac{5-\sqrt{37}}{2}\approx -0,541381265$$

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.

$$w^{2}-5w=3$$

Tenglamaning ikkala tarafidan 3 ni ayirish.

$$w^{2}-5w-3=3-3$$

O‘zidan 3 ayirilsa 0 qoladi.

$$w^{2}-5w-3=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 1 ni a, -5 ni b va -3 ni c bilan almashtiring.

$$w=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\left(-3\right)}}{2}$$

-5 kvadratini chiqarish.

$$w=\frac{-\left(-5\right)±\sqrt{25-4\left(-3\right)}}{2}$$

-4 ni -3 marotabaga ko'paytirish.

$$w=\frac{-\left(-5\right)±\sqrt{25+12}}{2}$$

25 ni 12 ga qo'shish.

$$w=\frac{-\left(-5\right)±\sqrt{37}}{2}$$

-5 ning teskarisi 5 ga teng.

$$w=\frac{5±\sqrt{37}}{2}$$

w=\frac{5±\sqrt{37}}{2} tenglamasini yeching, bunda ± musbat. 5 ni \sqrt{37} ga qo'shish.

$$w=\frac{\sqrt{37}+5}{2}$$

w=\frac{5±\sqrt{37}}{2} tenglamasini yeching, bunda ± manfiy. 5 dan \sqrt{37} ni ayirish.

$$w=\frac{5-\sqrt{37}}{2}$$

Tenglama yechildi.

$$w=\frac{\sqrt{37}+5}{2} w=\frac{5-\sqrt{37}}{2}$$

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.

$$w^{2}-5w=3$$

-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.

$$w^{2}-5w+\left(-\frac{5}{2}\right)^{2}=3+\left(-\frac{5}{2}\right)^{2}$$

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

$$w^{2}-5w+\frac{25}{4}=3+\frac{25}{4}$$

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

$$w^{2}-5w+\frac{25}{4}=\frac{37}{4}$$

w^{2}-5w+\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.

$$\left(w-\frac{5}{2}\right)^{2}=\frac{37}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(w-\frac{5}{2}\right)^{2}}=\sqrt{\frac{37}{4}}$$

Qisqartirish.

$$w-\frac{5}{2}=\frac{\sqrt{37}}{2} w-\frac{5}{2}=-\frac{\sqrt{37}}{2}$$

\frac{5}{2} ni tenglamaning ikkala tarafiga qo'shish.

$$w=\frac{\sqrt{37}+5}{2} w=\frac{5-\sqrt{37}}{2}$$