x^2+8x-2400=0 eching

$$x ^ { 2 } + 8 x - 2400 = 0$$

x uchun yechish

$$x=4\sqrt{151}-4\approx 45,15282291$$

$$x=-4\sqrt{151}-4\approx -53,15282291$$

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.

$$x^{2}+8x-2400=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 1 ni a, 8 ni b va -2400 ni c bilan almashtiring.

$$x=\frac{-8±\sqrt{8^{2}-4\left(-2400\right)}}{2}$$

8 kvadratini chiqarish.

$$x=\frac{-8±\sqrt{64-4\left(-2400\right)}}{2}$$

-4 ni -2400 marotabaga ko'paytirish.

$$x=\frac{-8±\sqrt{64+9600}}{2}$$

64 ni 9600 ga qo'shish.

$$x=\frac{-8±\sqrt{9664}}{2}$$

9664 ning kvadrat ildizini chiqarish.

$$x=\frac{-8±8\sqrt{151}}{2}$$

x=\frac{-8±8\sqrt{151}}{2} tenglamasini yeching, bunda ± musbat. -8 ni 8\sqrt{151} ga qo'shish.

$$x=\frac{8\sqrt{151}-8}{2}$$

-8+8\sqrt{151} ni 2 ga bo'lish.

$$x=4\sqrt{151}-4$$

x=\frac{-8±8\sqrt{151}}{2} tenglamasini yeching, bunda ± manfiy. -8 dan 8\sqrt{151} ni ayirish.

$$x=\frac{-8\sqrt{151}-8}{2}$$

-8-8\sqrt{151} ni 2 ga bo'lish.

$$x=-4\sqrt{151}-4$$

Tenglama yechildi.

$$x=4\sqrt{151}-4 x=-4\sqrt{151}-4$$

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.

$$x^{2}+8x-2400=0$$

2400 ni tenglamaning ikkala tarafiga qo'shish.

$$x^{2}+8x-2400-\left(-2400\right)=-\left(-2400\right)$$

O‘zidan -2400 ayirilsa 0 qoladi.

$$x^{2}+8x=-\left(-2400\right)$$

0 dan -2400 ni ayirish.

$$x^{2}+8x=2400$$

8 ni bo‘lish, x shartining koeffitsienti, 2 ga 4 olish uchun. Keyin, 4 ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+8x+4^{2}=2400+4^{2}$$

4 kvadratini chiqarish.

$$x^{2}+8x+16=2400+16$$

2400 ni 16 ga qo'shish.

$$x^{2}+8x+16=2416$$

x^{2}+8x+16 omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+4\right)^{2}=2416$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+4\right)^{2}}=\sqrt{2416}$$

Qisqartirish.

$$x+4=4\sqrt{151} x+4=-4\sqrt{151}$$

Tenglamaning ikkala tarafidan 4 ni ayirish.

$$x=4\sqrt{151}-4 x=-4\sqrt{151}-4$$