5x^2+10x-20=0 eching

$$5 { x }^{ 2 } +10x-20=0$$

x uchun yechish (complex solution)

$$x=\sqrt{5}-1\approx 1,236067977$$

$$x=-\left(\sqrt{5}+1\right)\approx -3,236067977$$

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.

$$5x^{2}+10x-20=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 5 ni a, 10 ni b va -20 ni c bilan almashtiring.

$$x=\frac{-10±\sqrt{10^{2}-4\times 5\left(-20\right)}}{2\times 5}$$

10 kvadratini chiqarish.

$$x=\frac{-10±\sqrt{100-4\times 5\left(-20\right)}}{2\times 5}$$

-4 ni 5 marotabaga ko'paytirish.

$$x=\frac{-10±\sqrt{100-20\left(-20\right)}}{2\times 5}$$

-20 ni -20 marotabaga ko'paytirish.

$$x=\frac{-10±\sqrt{100+400}}{2\times 5}$$

100 ni 400 ga qo'shish.

$$x=\frac{-10±\sqrt{500}}{2\times 5}$$

500 ning kvadrat ildizini chiqarish.

$$x=\frac{-10±10\sqrt{5}}{2\times 5}$$

2 ni 5 marotabaga ko'paytirish.

$$x=\frac{-10±10\sqrt{5}}{10}$$

x=\frac{-10±10\sqrt{5}}{10} tenglamasini yeching, bunda ± musbat. -10 ni 10\sqrt{5} ga qo'shish.

$$x=\frac{10\sqrt{5}-10}{10}$$

-10+10\sqrt{5} ni 10 ga bo'lish.

$$x=\sqrt{5}-1$$

x=\frac{-10±10\sqrt{5}}{10} tenglamasini yeching, bunda ± manfiy. -10 dan 10\sqrt{5} ni ayirish.

$$x=\frac{-10\sqrt{5}-10}{10}$$

-10-10\sqrt{5} ni 10 ga bo'lish.

$$x=-\sqrt{5}-1$$

Tenglama yechildi.

$$x=\sqrt{5}-1 x=-\sqrt{5}-1$$

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.

$$5x^{2}+10x-20=0$$

20 ni tenglamaning ikkala tarafiga qo'shish.

$$5x^{2}+10x-20-\left(-20\right)=-\left(-20\right)$$

O‘zidan -20 ayirilsa 0 qoladi.

$$5x^{2}+10x=-\left(-20\right)$$

0 dan -20 ni ayirish.

$$5x^{2}+10x=20$$

Ikki tarafini 5 ga bo‘ling.

$$\frac{5x^{2}+10x}{5}=\frac{20}{5}$$

5 ga bo'lish 5 ga ko'paytirishni bekor qiladi.

$$x^{2}+\frac{10}{5}x=\frac{20}{5}$$

10 ni 5 ga bo'lish.

$$x^{2}+2x=\frac{20}{5}$$

20 ni 5 ga bo'lish.

$$x^{2}+2x=4$$

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.

$$x^{2}+2x+1^{2}=4+1^{2}$$

1 kvadratini chiqarish.

$$x^{2}+2x+1=4+1$$

4 ni 1 ga qo'shish.

$$x^{2}+2x+1=5$$

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.

$$\left(x+1\right)^{2}=5$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{5}$$

Qisqartirish.

$$x+1=\sqrt{5} x+1=-\sqrt{5}$$

Tenglamaning ikkala tarafidan 1 ni ayirish.

$$x=\sqrt{5}-1 x=-\sqrt{5}-1$$

x uchun yechish

$$x=\sqrt{5}-1\approx 1,236067977$$

$$x=-\sqrt{5}-1\approx -3,236067977$$

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.

$$5x^{2}+10x-20=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 5 ni a, 10 ni b va -20 ni c bilan almashtiring.

$$x=\frac{-10±\sqrt{10^{2}-4\times 5\left(-20\right)}}{2\times 5}$$

10 kvadratini chiqarish.

$$x=\frac{-10±\sqrt{100-4\times 5\left(-20\right)}}{2\times 5}$$

-4 ni 5 marotabaga ko'paytirish.

$$x=\frac{-10±\sqrt{100-20\left(-20\right)}}{2\times 5}$$

-20 ni -20 marotabaga ko'paytirish.

$$x=\frac{-10±\sqrt{100+400}}{2\times 5}$$

100 ni 400 ga qo'shish.

$$x=\frac{-10±\sqrt{500}}{2\times 5}$$

500 ning kvadrat ildizini chiqarish.

$$x=\frac{-10±10\sqrt{5}}{2\times 5}$$

2 ni 5 marotabaga ko'paytirish.

$$x=\frac{-10±10\sqrt{5}}{10}$$

x=\frac{-10±10\sqrt{5}}{10} tenglamasini yeching, bunda ± musbat. -10 ni 10\sqrt{5} ga qo'shish.

$$x=\frac{10\sqrt{5}-10}{10}$$

-10+10\sqrt{5} ni 10 ga bo'lish.

$$x=\sqrt{5}-1$$

x=\frac{-10±10\sqrt{5}}{10} tenglamasini yeching, bunda ± manfiy. -10 dan 10\sqrt{5} ni ayirish.

$$x=\frac{-10\sqrt{5}-10}{10}$$

-10-10\sqrt{5} ni 10 ga bo'lish.

$$x=-\sqrt{5}-1$$

Tenglama yechildi.

$$x=\sqrt{5}-1 x=-\sqrt{5}-1$$

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.

$$5x^{2}+10x-20=0$$

20 ni tenglamaning ikkala tarafiga qo'shish.

$$5x^{2}+10x-20-\left(-20\right)=-\left(-20\right)$$

O‘zidan -20 ayirilsa 0 qoladi.

$$5x^{2}+10x=-\left(-20\right)$$

0 dan -20 ni ayirish.

$$5x^{2}+10x=20$$

Ikki tarafini 5 ga bo‘ling.

$$\frac{5x^{2}+10x}{5}=\frac{20}{5}$$

5 ga bo'lish 5 ga ko'paytirishni bekor qiladi.

$$x^{2}+\frac{10}{5}x=\frac{20}{5}$$

10 ni 5 ga bo'lish.

$$x^{2}+2x=\frac{20}{5}$$

20 ni 5 ga bo'lish.

$$x^{2}+2x=4$$

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.

$$x^{2}+2x+1^{2}=4+1^{2}$$

1 kvadratini chiqarish.

$$x^{2}+2x+1=4+1$$

4 ni 1 ga qo'shish.

$$x^{2}+2x+1=5$$

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.

$$\left(x+1\right)^{2}=5$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{5}$$

Qisqartirish.

$$x+1=\sqrt{5} x+1=-\sqrt{5}$$

Tenglamaning ikkala tarafidan 1 ni ayirish.

$$x=\sqrt{5}-1 x=-\sqrt{5}-1$$