z^2-3z+4/4=0 eching

$$z ^ { 2 } - 3 z + \frac { 4 } { 4 } = 0$$

z uchun yechish

$$z = \frac{\sqrt{5} + 3}{2} \approx 2,618033989$$

$$z=\frac{3-\sqrt{5}}{2}\approx 0,381966011$$

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.

$$z^{2}-3z+1=0$$

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

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

-3 kvadratini chiqarish.

$$z=\frac{-\left(-3\right)±\sqrt{9-4}}{2}$$

9 ni -4 ga qo'shish.

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

-3 ning teskarisi 3 ga teng.

$$z=\frac{3±\sqrt{5}}{2}$$

z=\frac{3±\sqrt{5}}{2} tenglamasini yeching, bunda ± musbat. 3 ni \sqrt{5} ga qo'shish.

$$z=\frac{\sqrt{5}+3}{2}$$

z=\frac{3±\sqrt{5}}{2} tenglamasini yeching, bunda ± manfiy. 3 dan \sqrt{5} ni ayirish.

$$z=\frac{3-\sqrt{5}}{2}$$

Tenglama yechildi.

$$z=\frac{\sqrt{5}+3}{2} z=\frac{3-\sqrt{5}}{2}$$

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.

$$z^{2}-3z+1=0$$

Tenglamaning ikkala tarafidan 1 ni ayirish.

$$z^{2}-3z+1-1=-1$$

O‘zidan 1 ayirilsa 0 qoladi.

$$z^{2}-3z=-1$$

-3 ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{3}{2} olish uchun. Keyin, -\frac{3}{2} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$z^{2}-3z+\left(-\frac{3}{2}\right)^{2}=-1+\left(-\frac{3}{2}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{3}{2} kvadratini chiqarish.

$$z^{2}-3z+\frac{9}{4}=-1+\frac{9}{4}$$

-1 ni \frac{9}{4} ga qo'shish.

$$z^{2}-3z+\frac{9}{4}=\frac{5}{4}$$

z^{2}-3z+\frac{9}{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.

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

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

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

Qisqartirish.

$$z-\frac{3}{2}=\frac{\sqrt{5}}{2} z-\frac{3}{2}=-\frac{\sqrt{5}}{2}$$

\frac{3}{2} ni tenglamaning ikkala tarafiga qo'shish.

$$z=\frac{\sqrt{5}+3}{2} z=\frac{3-\sqrt{5}}{2}$$