x^2-3x+20=50 eching

$$x ^ { 2 } - 3 x + 20 = 50$$

x uchun yechish

$$x = \frac{\sqrt{129} + 3}{2} \approx 7,178908346$$

$$x=\frac{3-\sqrt{129}}{2}\approx -4,178908346$$

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.

$$x^{2}-3x+20=50$$

Tenglamaning ikkala tarafidan 50 ni ayirish.

$$x^{2}-3x+20-50=50-50$$

O‘zidan 50 ayirilsa 0 qoladi.

$$x^{2}-3x+20-50=0$$

20 dan 50 ni ayirish.

$$x^{2}-3x-30=0$$

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 -30 ni c bilan almashtiring.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-30\right)}}{2}$$

-3 kvadratini chiqarish.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-30\right)}}{2}$$

-4 ni -30 marotabaga ko'paytirish.

$$x=\frac{-\left(-3\right)±\sqrt{9+120}}{2}$$

9 ni 120 ga qo'shish.

$$x=\frac{-\left(-3\right)±\sqrt{129}}{2}$$

-3 ning teskarisi 3 ga teng.

$$x=\frac{3±\sqrt{129}}{2}$$

x=\frac{3±\sqrt{129}}{2} tenglamasini yeching, bunda ± musbat. 3 ni \sqrt{129} ga qo'shish.

$$x=\frac{\sqrt{129}+3}{2}$$

x=\frac{3±\sqrt{129}}{2} tenglamasini yeching, bunda ± manfiy. 3 dan \sqrt{129} ni ayirish.

$$x=\frac{3-\sqrt{129}}{2}$$

Tenglama yechildi.

$$x=\frac{\sqrt{129}+3}{2} x=\frac{3-\sqrt{129}}{2}$$

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.

$$x^{2}-3x+20=50$$

Tenglamaning ikkala tarafidan 20 ni ayirish.

$$x^{2}-3x+20-20=50-20$$

O‘zidan 20 ayirilsa 0 qoladi.

$$x^{2}-3x=50-20$$

50 dan 20 ni ayirish.

$$x^{2}-3x=30$$

-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.

$$x^{2}-3x+\left(-\frac{3}{2}\right)^{2}=30+\left(-\frac{3}{2}\right)^{2}$$

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

$$x^{2}-3x+\frac{9}{4}=30+\frac{9}{4}$$

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

$$x^{2}-3x+\frac{9}{4}=\frac{129}{4}$$

x^{2}-3x+\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.

$$\left(x-\frac{3}{2}\right)^{2}=\frac{129}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{129}{4}}$$

Qisqartirish.

$$x-\frac{3}{2}=\frac{\sqrt{129}}{2} x-\frac{3}{2}=-\frac{\sqrt{129}}{2}$$

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

$$x=\frac{\sqrt{129}+3}{2} x=\frac{3-\sqrt{129}}{2}$$