8x^2-5x+87=7x^2+5x eching

$$8 x ^ { 2 } - 5 x + 87 = 7 x ^ { 2 } + 5 x$$

x uchun yechish (complex solution)

$$x=5+\sqrt{62}i\approx 5+7,874007874i$$

$$x=-\sqrt{62}i+5\approx 5-7,874007874i$$

Kvadrat tenglamalaridan foydalanish qoidalari

Ikkala tarafdan 7x^{2} ni ayirish.

$$8x^{2}-5x+87-7x^{2}=5x$$

x^{2} ni olish uchun 8x^{2} va -7x^{2} ni birlashtirish.

$$x^{2}-5x+87=5x$$

Ikkala tarafdan 5x ni ayirish.

$$x^{2}-5x+87-5x=0$$

-10x ni olish uchun -5x va -5x ni birlashtirish.

$$x^{2}-10x+87=0$$

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 87 ni c bilan almashtiring.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 87}}{2}$$

-10 kvadratini chiqarish.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\times 87}}{2}$$

-4 ni 87 marotabaga ko'paytirish.

$$x=\frac{-\left(-10\right)±\sqrt{100-348}}{2}$$

100 ni -348 ga qo'shish.

$$x=\frac{-\left(-10\right)±\sqrt{-248}}{2}$$

-248 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-10\right)±2\sqrt{62}i}{2}$$

-10 ning teskarisi 10 ga teng.

$$x=\frac{10±2\sqrt{62}i}{2}$$

x=\frac{10±2\sqrt{62}i}{2} tenglamasini yeching, bunda ± musbat. 10 ni 2i\sqrt{62} ga qo'shish.

$$x=\frac{10+2\sqrt{62}i}{2}$$

10+2i\sqrt{62} ni 2 ga bo'lish.

$$x=5+\sqrt{62}i$$

x=\frac{10±2\sqrt{62}i}{2} tenglamasini yeching, bunda ± manfiy. 10 dan 2i\sqrt{62} ni ayirish.

$$x=\frac{-2\sqrt{62}i+10}{2}$$

10-2i\sqrt{62} ni 2 ga bo'lish.

$$x=-\sqrt{62}i+5$$

Tenglama yechildi.

$$x=5+\sqrt{62}i x=-\sqrt{62}i+5$$

Kvadratni yakunlash bo'yicha qadamlar

Ikkala tarafdan 7x^{2} ni ayirish.

$$8x^{2}-5x+87-7x^{2}=5x$$

x^{2} ni olish uchun 8x^{2} va -7x^{2} ni birlashtirish.

$$x^{2}-5x+87=5x$$

Ikkala tarafdan 5x ni ayirish.

$$x^{2}-5x+87-5x=0$$

-10x ni olish uchun -5x va -5x ni birlashtirish.

$$x^{2}-10x+87=0$$

Ikkala tarafdan 87 ni ayirish. Har qanday sonni noldan ayirsangiz, o‘zining manfiyi chiqadi.

$$x^{2}-10x=-87$$

-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.

$$x^{2}-10x+\left(-5\right)^{2}=-87+\left(-5\right)^{2}$$

-5 kvadratini chiqarish.

$$x^{2}-10x+25=-87+25$$

-87 ni 25 ga qo'shish.

$$x^{2}-10x+25=-62$$

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.

$$\left(x-5\right)^{2}=-62$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-5\right)^{2}}=\sqrt{-62}$$

Qisqartirish.

$$x-5=\sqrt{62}i x-5=-\sqrt{62}i$$

5 ni tenglamaning ikkala tarafiga qo'shish.

$$x=5+\sqrt{62}i x=-\sqrt{62}i+5$$