x^2+10x+25=7 eching

$$x ^ { 2 } + 10 x + 25 = 7$$

x uchun yechish (complex solution)

$$x=\sqrt{7}-5\approx -2,354248689$$

$$x=-\left(\sqrt{7}+5\right)\approx -7,645751311$$

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.

$$x^{2}+10x+25=7$$

Tenglamaning ikkala tarafidan 7 ni ayirish.

$$x^{2}+10x+25-7=7-7$$

O‘zidan 7 ayirilsa 0 qoladi.

$$x^{2}+10x+25-7=0$$

25 dan 7 ni ayirish.

$$x^{2}+10x+18=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 1 ni a, 10 ni b va 18 ni c bilan almashtiring.

$$x=\frac{-10±\sqrt{10^{2}-4\times 18}}{2}$$

10 kvadratini chiqarish.

$$x=\frac{-10±\sqrt{100-4\times 18}}{2}$$

-4 ni 18 marotabaga ko'paytirish.

$$x=\frac{-10±\sqrt{100-72}}{2}$$

100 ni -72 ga qo'shish.

$$x=\frac{-10±\sqrt{28}}{2}$$

28 ning kvadrat ildizini chiqarish.

$$x=\frac{-10±2\sqrt{7}}{2}$$

x=\frac{-10±2\sqrt{7}}{2} tenglamasini yeching, bunda ± musbat. -10 ni 2\sqrt{7} ga qo'shish.

$$x=\frac{2\sqrt{7}-10}{2}$$

-10+2\sqrt{7} ni 2 ga bo'lish.

$$x=\sqrt{7}-5$$

x=\frac{-10±2\sqrt{7}}{2} tenglamasini yeching, bunda ± manfiy. -10 dan 2\sqrt{7} ni ayirish.

$$x=\frac{-2\sqrt{7}-10}{2}$$

-10-2\sqrt{7} ni 2 ga bo'lish.

$$x=-\sqrt{7}-5$$

Tenglama yechildi.

$$x=\sqrt{7}-5 x=-\sqrt{7}-5$$

Kvadratni yakunlash bo'yicha qadamlar

x^{2}+10x+25 omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+5\right)^{2}=7$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+5\right)^{2}}=\sqrt{7}$$

Qisqartirish.

$$x+5=\sqrt{7} x+5=-\sqrt{7}$$

Tenglamaning ikkala tarafidan 5 ni ayirish.

$$x=\sqrt{7}-5 x=-\sqrt{7}-5$$

x uchun yechish

$$x=\sqrt{7}-5\approx -2,354248689$$

$$x=-\sqrt{7}-5\approx -7,645751311$$

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.

$$x^{2}+10x+25=7$$

Tenglamaning ikkala tarafidan 7 ni ayirish.

$$x^{2}+10x+25-7=7-7$$

O‘zidan 7 ayirilsa 0 qoladi.

$$x^{2}+10x+25-7=0$$

25 dan 7 ni ayirish.

$$x^{2}+10x+18=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 1 ni a, 10 ni b va 18 ni c bilan almashtiring.

$$x=\frac{-10±\sqrt{10^{2}-4\times 18}}{2}$$

10 kvadratini chiqarish.

$$x=\frac{-10±\sqrt{100-4\times 18}}{2}$$

-4 ni 18 marotabaga ko'paytirish.

$$x=\frac{-10±\sqrt{100-72}}{2}$$

100 ni -72 ga qo'shish.

$$x=\frac{-10±\sqrt{28}}{2}$$

28 ning kvadrat ildizini chiqarish.

$$x=\frac{-10±2\sqrt{7}}{2}$$

x=\frac{-10±2\sqrt{7}}{2} tenglamasini yeching, bunda ± musbat. -10 ni 2\sqrt{7} ga qo'shish.

$$x=\frac{2\sqrt{7}-10}{2}$$

-10+2\sqrt{7} ni 2 ga bo'lish.

$$x=\sqrt{7}-5$$

x=\frac{-10±2\sqrt{7}}{2} tenglamasini yeching, bunda ± manfiy. -10 dan 2\sqrt{7} ni ayirish.

$$x=\frac{-2\sqrt{7}-10}{2}$$

-10-2\sqrt{7} ni 2 ga bo'lish.

$$x=-\sqrt{7}-5$$

Tenglama yechildi.

$$x=\sqrt{7}-5 x=-\sqrt{7}-5$$

Kvadratni yakunlash bo'yicha qadamlar

x^{2}+10x+25 omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+5\right)^{2}=7$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+5\right)^{2}}=\sqrt{7}$$

Qisqartirish.

$$x+5=\sqrt{7} x+5=-\sqrt{7}$$

Tenglamaning ikkala tarafidan 5 ni ayirish.

$$x=\sqrt{7}-5 x=-\sqrt{7}-5$$