y^2-4y=6 eching

$$y ^ { 2 } - 4 y = 6$$

y uchun yechish

$$y=\sqrt{10}+2\approx 5,16227766$$

$$y=2-\sqrt{10}\approx -1,16227766$$

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.

$$y^{2}-4y=6$$

Tenglamaning ikkala tarafidan 6 ni ayirish.

$$y^{2}-4y-6=6-6$$

O‘zidan 6 ayirilsa 0 qoladi.

$$y^{2}-4y-6=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 1 ni a, -4 ni b va -6 ni c bilan almashtiring.

$$y=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\left(-6\right)}}{2}$$

-4 kvadratini chiqarish.

$$y=\frac{-\left(-4\right)±\sqrt{16-4\left(-6\right)}}{2}$$

-4 ni -6 marotabaga ko'paytirish.

$$y=\frac{-\left(-4\right)±\sqrt{16+24}}{2}$$

16 ni 24 ga qo'shish.

$$y=\frac{-\left(-4\right)±\sqrt{40}}{2}$$

40 ning kvadrat ildizini chiqarish.

$$y=\frac{-\left(-4\right)±2\sqrt{10}}{2}$$

-4 ning teskarisi 4 ga teng.

$$y=\frac{4±2\sqrt{10}}{2}$$

y=\frac{4±2\sqrt{10}}{2} tenglamasini yeching, bunda ± musbat. 4 ni 2\sqrt{10} ga qo'shish.

$$y=\frac{2\sqrt{10}+4}{2}$$

4+2\sqrt{10} ni 2 ga bo'lish.

$$y=\sqrt{10}+2$$

y=\frac{4±2\sqrt{10}}{2} tenglamasini yeching, bunda ± manfiy. 4 dan 2\sqrt{10} ni ayirish.

$$y=\frac{4-2\sqrt{10}}{2}$$

4-2\sqrt{10} ni 2 ga bo'lish.

$$y=2-\sqrt{10}$$

Tenglama yechildi.

$$y=\sqrt{10}+2 y=2-\sqrt{10}$$

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.

$$y^{2}-4y=6$$

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

$$y^{2}-4y+\left(-2\right)^{2}=6+\left(-2\right)^{2}$$

-2 kvadratini chiqarish.

$$y^{2}-4y+4=6+4$$

6 ni 4 ga qo'shish.

$$y^{2}-4y+4=10$$

y^{2}-4y+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.

$$\left(y-2\right)^{2}=10$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(y-2\right)^{2}}=\sqrt{10}$$

Qisqartirish.

$$y-2=\sqrt{10} y-2=-\sqrt{10}$$

2 ni tenglamaning ikkala tarafiga qo'shish.

$$y=\sqrt{10}+2 y=2-\sqrt{10}$$