x^2+30x=225-(20) eching

$$x ^ { 2 } + 30 x = 225 - ( 20 )$$

x uchun yechish (complex solution)

$$x=\sqrt{430}-15\approx 5,736441353$$

$$x=-\left(\sqrt{430}+15\right)\approx -35,736441353$$

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.

$$x^{2}+30x=205$$

Tenglamaning ikkala tarafidan 205 ni ayirish.

$$x^{2}+30x-205=205-205$$

O‘zidan 205 ayirilsa 0 qoladi.

$$x^{2}+30x-205=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 1 ni a, 30 ni b va -205 ni c bilan almashtiring.

$$x=\frac{-30±\sqrt{30^{2}-4\left(-205\right)}}{2}$$

30 kvadratini chiqarish.

$$x=\frac{-30±\sqrt{900-4\left(-205\right)}}{2}$$

-4 ni -205 marotabaga ko'paytirish.

$$x=\frac{-30±\sqrt{900+820}}{2}$$

900 ni 820 ga qo'shish.

$$x=\frac{-30±\sqrt{1720}}{2}$$

1720 ning kvadrat ildizini chiqarish.

$$x=\frac{-30±2\sqrt{430}}{2}$$

x=\frac{-30±2\sqrt{430}}{2} tenglamasini yeching, bunda ± musbat. -30 ni 2\sqrt{430} ga qo'shish.

$$x=\frac{2\sqrt{430}-30}{2}$$

-30+2\sqrt{430} ni 2 ga bo'lish.

$$x=\sqrt{430}-15$$

x=\frac{-30±2\sqrt{430}}{2} tenglamasini yeching, bunda ± manfiy. -30 dan 2\sqrt{430} ni ayirish.

$$x=\frac{-2\sqrt{430}-30}{2}$$

-30-2\sqrt{430} ni 2 ga bo'lish.

$$x=-\sqrt{430}-15$$

Tenglama yechildi.

$$x=\sqrt{430}-15 x=-\sqrt{430}-15$$

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.

$$x^{2}+30x=205$$

30 ni bo‘lish, x shartining koeffitsienti, 2 ga 15 olish uchun. Keyin, 15 ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+30x+15^{2}=205+15^{2}$$

15 kvadratini chiqarish.

$$x^{2}+30x+225=205+225$$

205 ni 225 ga qo'shish.

$$x^{2}+30x+225=430$$

x^{2}+30x+225 omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+15\right)^{2}=430$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+15\right)^{2}}=\sqrt{430}$$

Qisqartirish.

$$x+15=\sqrt{430} x+15=-\sqrt{430}$$

Tenglamaning ikkala tarafidan 15 ni ayirish.

$$x=\sqrt{430}-15 x=-\sqrt{430}-15$$

x uchun yechish

$$x=\sqrt{430}-15\approx 5,736441353$$

$$x=-\sqrt{430}-15\approx -35,736441353$$

Kvadrat tenglamalaridan foydalanish qoidalari

205 olish uchun 225 dan 20 ni ayirish.

$$x^{2}+30x=205$$

Ikkala tarafdan 205 ni ayirish.

$$x^{2}+30x-205=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 1 ni a, 30 ni b va -205 ni c bilan almashtiring.

$$x=\frac{-30±\sqrt{30^{2}-4\left(-205\right)}}{2}$$

30 kvadratini chiqarish.

$$x=\frac{-30±\sqrt{900-4\left(-205\right)}}{2}$$

-4 ni -205 marotabaga ko'paytirish.

$$x=\frac{-30±\sqrt{900+820}}{2}$$

900 ni 820 ga qo'shish.

$$x=\frac{-30±\sqrt{1720}}{2}$$

1720 ning kvadrat ildizini chiqarish.

$$x=\frac{-30±2\sqrt{430}}{2}$$

x=\frac{-30±2\sqrt{430}}{2} tenglamasini yeching, bunda ± musbat. -30 ni 2\sqrt{430} ga qo'shish.

$$x=\frac{2\sqrt{430}-30}{2}$$

-30+2\sqrt{430} ni 2 ga bo'lish.

$$x=\sqrt{430}-15$$

x=\frac{-30±2\sqrt{430}}{2} tenglamasini yeching, bunda ± manfiy. -30 dan 2\sqrt{430} ni ayirish.

$$x=\frac{-2\sqrt{430}-30}{2}$$

-30-2\sqrt{430} ni 2 ga bo'lish.

$$x=-\sqrt{430}-15$$

Tenglama yechildi.

$$x=\sqrt{430}-15 x=-\sqrt{430}-15$$

Kvadratni yakunlash bo'yicha qadamlar

205 olish uchun 225 dan 20 ni ayirish.

$$x^{2}+30x=205$$

30 ni bo‘lish, x shartining koeffitsienti, 2 ga 15 olish uchun. Keyin, 15 ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+30x+15^{2}=205+15^{2}$$

15 kvadratini chiqarish.

$$x^{2}+30x+225=205+225$$

205 ni 225 ga qo'shish.

$$x^{2}+30x+225=430$$

x^{2}+30x+225 omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+15\right)^{2}=430$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+15\right)^{2}}=\sqrt{430}$$

Qisqartirish.

$$x+15=\sqrt{430} x+15=-\sqrt{430}$$

Tenglamaning ikkala tarafidan 15 ni ayirish.

$$x=\sqrt{430}-15 x=-\sqrt{430}-15$$