x^2-10x=-39 eching

$$x ^ { 2 } - 10 x = - 39$$

x uchun yechish (complex solution)

$$x=5+\sqrt{14}i\approx 5+3,741657387i$$

$$x=-\sqrt{14}i+5\approx 5-3,741657387i$$

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.

$$x^{2}-10x=-39$$

39 ni tenglamaning ikkala tarafiga qo'shish.

$$x^{2}-10x-\left(-39\right)=-39-\left(-39\right)$$

O‘zidan -39 ayirilsa 0 qoladi.

$$x^{2}-10x-\left(-39\right)=0$$

0 dan -39 ni ayirish.

$$x^{2}-10x+39=0$$

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 39 ni c bilan almashtiring.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 39}}{2}$$

-10 kvadratini chiqarish.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\times 39}}{2}$$

-4 ni 39 marotabaga ko'paytirish.

$$x=\frac{-\left(-10\right)±\sqrt{100-156}}{2}$$

100 ni -156 ga qo'shish.

$$x=\frac{-\left(-10\right)±\sqrt{-56}}{2}$$

-56 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-10\right)±2\sqrt{14}i}{2}$$

-10 ning teskarisi 10 ga teng.

$$x=\frac{10±2\sqrt{14}i}{2}$$

x=\frac{10±2\sqrt{14}i}{2} tenglamasini yeching, bunda ± musbat. 10 ni 2i\sqrt{14} ga qo'shish.

$$x=\frac{10+2\sqrt{14}i}{2}$$

10+2i\sqrt{14} ni 2 ga bo'lish.

$$x=5+\sqrt{14}i$$

x=\frac{10±2\sqrt{14}i}{2} tenglamasini yeching, bunda ± manfiy. 10 dan 2i\sqrt{14} ni ayirish.

$$x=\frac{-2\sqrt{14}i+10}{2}$$

10-2i\sqrt{14} ni 2 ga bo'lish.

$$x=-\sqrt{14}i+5$$

Tenglama yechildi.

$$x=5+\sqrt{14}i x=-\sqrt{14}i+5$$

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.

$$x^{2}-10x=-39$$

-10 ni bo‘lish, x shartining koeffitsienti, 2 ga -5 olish uchun. Keyin, -5 ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}-10x+\left(-5\right)^{2}=-39+\left(-5\right)^{2}$$

-5 kvadratini chiqarish.

$$x^{2}-10x+25=-39+25$$

-39 ni 25 ga qo'shish.

$$x^{2}-10x+25=-14$$

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.

$$\left(x-5\right)^{2}=-14$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-5\right)^{2}}=\sqrt{-14}$$

Qisqartirish.

$$x-5=\sqrt{14}i x-5=-\sqrt{14}i$$

5 ni tenglamaning ikkala tarafiga qo'shish.

$$x=5+\sqrt{14}i x=-\sqrt{14}i+5$$