2x+x^2+6-10+(8*5)=52 eching

$$2x+ { x }^{ 2 } +6-10+(8 \times 5)=52$$

x uchun yechish (complex solution)

$$x=\sqrt{17}-1\approx 3,123105626$$

$$x=-\left(\sqrt{17}+1\right)\approx -5,123105626$$

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.

$$x^{2}+2x+36=52$$

Tenglamaning ikkala tarafidan 52 ni ayirish.

$$x^{2}+2x+36-52=52-52$$

O‘zidan 52 ayirilsa 0 qoladi.

$$x^{2}+2x+36-52=0$$

36 dan 52 ni ayirish.

$$x^{2}+2x-16=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 1 ni a, 2 ni b va -16 ni c bilan almashtiring.

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

2 kvadratini chiqarish.

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

-4 ni -16 marotabaga ko'paytirish.

$$x=\frac{-2±\sqrt{4+64}}{2}$$

4 ni 64 ga qo'shish.

$$x=\frac{-2±\sqrt{68}}{2}$$

68 ning kvadrat ildizini chiqarish.

$$x=\frac{-2±2\sqrt{17}}{2}$$

x=\frac{-2±2\sqrt{17}}{2} tenglamasini yeching, bunda ± musbat. -2 ni 2\sqrt{17} ga qo'shish.

$$x=\frac{2\sqrt{17}-2}{2}$$

-2+2\sqrt{17} ni 2 ga bo'lish.

$$x=\sqrt{17}-1$$

x=\frac{-2±2\sqrt{17}}{2} tenglamasini yeching, bunda ± manfiy. -2 dan 2\sqrt{17} ni ayirish.

$$x=\frac{-2\sqrt{17}-2}{2}$$

-2-2\sqrt{17} ni 2 ga bo'lish.

$$x=-\sqrt{17}-1$$

Tenglama yechildi.

$$x=\sqrt{17}-1 x=-\sqrt{17}-1$$

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.

$$x^{2}+2x+36=52$$

Tenglamaning ikkala tarafidan 36 ni ayirish.

$$x^{2}+2x+36-36=52-36$$

O‘zidan 36 ayirilsa 0 qoladi.

$$x^{2}+2x=52-36$$

52 dan 36 ni ayirish.

$$x^{2}+2x=16$$

2 ni bo‘lish, x shartining koeffitsienti, 2 ga 1 olish uchun. Keyin, 1 ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+2x+1^{2}=16+1^{2}$$

1 kvadratini chiqarish.

$$x^{2}+2x+1=16+1$$

16 ni 1 ga qo'shish.

$$x^{2}+2x+1=17$$

x^{2}+2x+1 omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+1\right)^{2}=17$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{17}$$

Qisqartirish.

$$x+1=\sqrt{17} x+1=-\sqrt{17}$$

Tenglamaning ikkala tarafidan 1 ni ayirish.

$$x=\sqrt{17}-1 x=-\sqrt{17}-1$$

x uchun yechish

$$x=\sqrt{17}-1\approx 3,123105626$$

$$x=-\sqrt{17}-1\approx -5,123105626$$

Kvadrat tenglamalaridan foydalanish qoidalari

-4 olish uchun 6 dan 10 ni ayirish.

$$2x+x^{2}-4+8\times 5=52$$

40 hosil qilish uchun 8 va 5 ni ko'paytirish.

$$2x+x^{2}-4+40=52$$

36 olish uchun -4 va 40'ni qo'shing.

$$2x+x^{2}+36=52$$

Ikkala tarafdan 52 ni ayirish.

$$2x+x^{2}+36-52=0$$

-16 olish uchun 36 dan 52 ni ayirish.

$$2x+x^{2}-16=0$$

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.

$$x^{2}+2x-16=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 1 ni a, 2 ni b va -16 ni c bilan almashtiring.

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

2 kvadratini chiqarish.

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

-4 ni -16 marotabaga ko'paytirish.

$$x=\frac{-2±\sqrt{4+64}}{2}$$

4 ni 64 ga qo'shish.

$$x=\frac{-2±\sqrt{68}}{2}$$

68 ning kvadrat ildizini chiqarish.

$$x=\frac{-2±2\sqrt{17}}{2}$$

x=\frac{-2±2\sqrt{17}}{2} tenglamasini yeching, bunda ± musbat. -2 ni 2\sqrt{17} ga qo'shish.

$$x=\frac{2\sqrt{17}-2}{2}$$

-2+2\sqrt{17} ni 2 ga bo'lish.

$$x=\sqrt{17}-1$$

x=\frac{-2±2\sqrt{17}}{2} tenglamasini yeching, bunda ± manfiy. -2 dan 2\sqrt{17} ni ayirish.

$$x=\frac{-2\sqrt{17}-2}{2}$$

-2-2\sqrt{17} ni 2 ga bo'lish.

$$x=-\sqrt{17}-1$$

Tenglama yechildi.

$$x=\sqrt{17}-1 x=-\sqrt{17}-1$$

Kvadratni yakunlash bo'yicha qadamlar

-4 olish uchun 6 dan 10 ni ayirish.

$$2x+x^{2}-4+8\times 5=52$$

40 hosil qilish uchun 8 va 5 ni ko'paytirish.

$$2x+x^{2}-4+40=52$$

36 olish uchun -4 va 40'ni qo'shing.

$$2x+x^{2}+36=52$$

Ikkala tarafdan 36 ni ayirish.

$$2x+x^{2}=52-36$$

16 olish uchun 52 dan 36 ni ayirish.

$$2x+x^{2}=16$$

Bu kabi kvadrat tenglamalarni kvadratni yakunlab yechish mumkin. Kvadratni yechish uchun tenglama avval ushbu shaklda bo'lishi shart: x^{2}+bx=c.

$$x^{2}+2x=16$$

2 ni bo‘lish, x shartining koeffitsienti, 2 ga 1 olish uchun. Keyin, 1 ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+2x+1^{2}=16+1^{2}$$

1 kvadratini chiqarish.

$$x^{2}+2x+1=16+1$$

16 ni 1 ga qo'shish.

$$x^{2}+2x+1=17$$

x^{2}+2x+1 omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+1\right)^{2}=17$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{17}$$

Qisqartirish.

$$x+1=\sqrt{17} x+1=-\sqrt{17}$$

Tenglamaning ikkala tarafidan 1 ni ayirish.

$$x=\sqrt{17}-1 x=-\sqrt{17}-1$$