x^2-10x+90=0 eching

$$x ^ { 2 } - 10 x + 90 = 0$$

x uchun yechish (complex solution)

$$x=5+\sqrt{65}i\approx 5+8,062257748i$$

$$x=-\sqrt{65}i+5\approx 5-8,062257748i$$

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.

$$x^{2}-10x+90=0$$

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 90 ni c bilan almashtiring.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 90}}{2}$$

-10 kvadratini chiqarish.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\times 90}}{2}$$

-4 ni 90 marotabaga ko'paytirish.

$$x=\frac{-\left(-10\right)±\sqrt{100-360}}{2}$$

100 ni -360 ga qo'shish.

$$x=\frac{-\left(-10\right)±\sqrt{-260}}{2}$$

-260 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-10\right)±2\sqrt{65}i}{2}$$

-10 ning teskarisi 10 ga teng.

$$x=\frac{10±2\sqrt{65}i}{2}$$

x=\frac{10±2\sqrt{65}i}{2} tenglamasini yeching, bunda ± musbat. 10 ni 2i\sqrt{65} ga qo'shish.

$$x=\frac{10+2\sqrt{65}i}{2}$$

10+2i\sqrt{65} ni 2 ga bo'lish.

$$x=5+\sqrt{65}i$$

x=\frac{10±2\sqrt{65}i}{2} tenglamasini yeching, bunda ± manfiy. 10 dan 2i\sqrt{65} ni ayirish.

$$x=\frac{-2\sqrt{65}i+10}{2}$$

10-2i\sqrt{65} ni 2 ga bo'lish.

$$x=-\sqrt{65}i+5$$

Tenglama yechildi.

$$x=5+\sqrt{65}i x=-\sqrt{65}i+5$$

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.

$$x^{2}-10x+90=0$$

Tenglamaning ikkala tarafidan 90 ni ayirish.

$$x^{2}-10x+90-90=-90$$

O‘zidan 90 ayirilsa 0 qoladi.

$$x^{2}-10x=-90$$

-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.

$$x^{2}-10x+\left(-5\right)^{2}=-90+\left(-5\right)^{2}$$

-5 kvadratini chiqarish.

$$x^{2}-10x+25=-90+25$$

-90 ni 25 ga qo'shish.

$$x^{2}-10x+25=-65$$

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.

$$\left(x-5\right)^{2}=-65$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-5\right)^{2}}=\sqrt{-65}$$

Qisqartirish.

$$x-5=\sqrt{65}i x-5=-\sqrt{65}i$$

5 ni tenglamaning ikkala tarafiga qo'shish.

$$x=5+\sqrt{65}i x=-\sqrt{65}i+5$$