3x+6-x^2=5 eching

$$3 x + 6 - x ^ { 2 } = 5$$

x uchun yechish

$$x = \frac{\sqrt{13} + 3}{2} \approx 3,302775638$$

$$x=\frac{3-\sqrt{13}}{2}\approx -0,302775638$$

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.

$$-x^{2}+3x+6=5$$

Tenglamaning ikkala tarafidan 5 ni ayirish.

$$-x^{2}+3x+6-5=5-5$$

O‘zidan 5 ayirilsa 0 qoladi.

$$-x^{2}+3x+6-5=0$$

6 dan 5 ni ayirish.

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

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.

$$x=\frac{-3±\sqrt{3^{2}-4\left(-1\right)}}{2\left(-1\right)}$$

3 kvadratini chiqarish.

$$x=\frac{-3±\sqrt{9-4\left(-1\right)}}{2\left(-1\right)}$$

-4 ni -1 marotabaga ko'paytirish.

$$x=\frac{-3±\sqrt{9+4}}{2\left(-1\right)}$$

9 ni 4 ga qo'shish.

$$x=\frac{-3±\sqrt{13}}{2\left(-1\right)}$$

2 ni -1 marotabaga ko'paytirish.

$$x=\frac{-3±\sqrt{13}}{-2}$$

x=\frac{-3±\sqrt{13}}{-2} tenglamasini yeching, bunda ± musbat. -3 ni \sqrt{13} ga qo'shish.

$$x=\frac{\sqrt{13}-3}{-2}$$

-3+\sqrt{13} ni -2 ga bo'lish.

$$x=\frac{3-\sqrt{13}}{2}$$

x=\frac{-3±\sqrt{13}}{-2} tenglamasini yeching, bunda ± manfiy. -3 dan \sqrt{13} ni ayirish.

$$x=\frac{-\sqrt{13}-3}{-2}$$

-3-\sqrt{13} ni -2 ga bo'lish.

$$x=\frac{\sqrt{13}+3}{2}$$

Tenglama yechildi.

$$x=\frac{3-\sqrt{13}}{2} x=\frac{\sqrt{13}+3}{2}$$

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.

$$-x^{2}+3x+6=5$$

Tenglamaning ikkala tarafidan 6 ni ayirish.

$$-x^{2}+3x+6-6=5-6$$

O‘zidan 6 ayirilsa 0 qoladi.

$$-x^{2}+3x=5-6$$

5 dan 6 ni ayirish.

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

Ikki tarafini -1 ga bo‘ling.

$$\frac{-x^{2}+3x}{-1}=-\frac{1}{-1}$$

-1 ga bo'lish -1 ga ko'paytirishni bekor qiladi.

$$x^{2}+\frac{3}{-1}x=-\frac{1}{-1}$$

3 ni -1 ga bo'lish.

$$x^{2}-3x=-\frac{1}{-1}$$

-1 ni -1 ga bo'lish.

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

-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.

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

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

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

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

$$x^{2}-3x+\frac{9}{4}=\frac{13}{4}$$

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.

$$\left(x-\frac{3}{2}\right)^{2}=\frac{13}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{13}{4}}$$

Qisqartirish.

$$x-\frac{3}{2}=\frac{\sqrt{13}}{2} x-\frac{3}{2}=-\frac{\sqrt{13}}{2}$$

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

$$x=\frac{\sqrt{13}+3}{2} x=\frac{3-\sqrt{13}}{2}$$