x^2-8x+10=13x eching

$$x ^ { 2 } - 8 x + 10 = 13 x$$

x uchun yechish

$$x = \frac{\sqrt{401} + 21}{2} \approx 20,512492197$$

$$x=\frac{21-\sqrt{401}}{2}\approx 0,487507803$$

Kvadrat tenglamalaridan foydalanish qoidalari

Ikkala tarafdan 13x ni ayirish.

$$x^{2}-8x+10-13x=0$$

-21x ni olish uchun -8x va -13x ni birlashtirish.

$$x^{2}-21x+10=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 1 ni a, -21 ni b va 10 ni c bilan almashtiring.

$$x=\frac{-\left(-21\right)±\sqrt{\left(-21\right)^{2}-4\times 10}}{2}$$

-21 kvadratini chiqarish.

$$x=\frac{-\left(-21\right)±\sqrt{441-4\times 10}}{2}$$

-4 ni 10 marotabaga ko'paytirish.

$$x=\frac{-\left(-21\right)±\sqrt{441-40}}{2}$$

441 ni -40 ga qo'shish.

$$x=\frac{-\left(-21\right)±\sqrt{401}}{2}$$

-21 ning teskarisi 21 ga teng.

$$x=\frac{21±\sqrt{401}}{2}$$

x=\frac{21±\sqrt{401}}{2} tenglamasini yeching, bunda ± musbat. 21 ni \sqrt{401} ga qo'shish.

$$x=\frac{\sqrt{401}+21}{2}$$

x=\frac{21±\sqrt{401}}{2} tenglamasini yeching, bunda ± manfiy. 21 dan \sqrt{401} ni ayirish.

$$x=\frac{21-\sqrt{401}}{2}$$

Tenglama yechildi.

$$x=\frac{\sqrt{401}+21}{2} x=\frac{21-\sqrt{401}}{2}$$

Kvadratni yakunlash bo'yicha qadamlar

Ikkala tarafdan 13x ni ayirish.

$$x^{2}-8x+10-13x=0$$

-21x ni olish uchun -8x va -13x ni birlashtirish.

$$x^{2}-21x+10=0$$

Ikkala tarafdan 10 ni ayirish. Har qanday sonni noldan ayirsangiz, o‘zining manfiyi chiqadi.

$$x^{2}-21x=-10$$

-21 ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{21}{2} olish uchun. Keyin, -\frac{21}{2} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}-21x+\left(-\frac{21}{2}\right)^{2}=-10+\left(-\frac{21}{2}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{21}{2} kvadratini chiqarish.

$$x^{2}-21x+\frac{441}{4}=-10+\frac{441}{4}$$

-10 ni \frac{441}{4} ga qo'shish.

$$x^{2}-21x+\frac{441}{4}=\frac{401}{4}$$

x^{2}-21x+\frac{441}{4} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x-\frac{21}{2}\right)^{2}=\frac{401}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{21}{2}\right)^{2}}=\sqrt{\frac{401}{4}}$$

Qisqartirish.

$$x-\frac{21}{2}=\frac{\sqrt{401}}{2} x-\frac{21}{2}=-\frac{\sqrt{401}}{2}$$

\frac{21}{2} ni tenglamaning ikkala tarafiga qo'shish.

$$x=\frac{\sqrt{401}+21}{2} x=\frac{21-\sqrt{401}}{2}$$