x^2-10x-400=0 eching

$${ x }^{ 2 } -10x-400=0$$

x uchun yechish

$$x=5\sqrt{17}+5\approx 25,615528128$$

$$x=5-5\sqrt{17}\approx -15,615528128$$

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.

$$x^{2}-10x-400=0$$

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 -400 ni c bilan almashtiring.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\left(-400\right)}}{2}$$

-10 kvadratini chiqarish.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\left(-400\right)}}{2}$$

-4 ni -400 marotabaga ko'paytirish.

$$x=\frac{-\left(-10\right)±\sqrt{100+1600}}{2}$$

100 ni 1600 ga qo'shish.

$$x=\frac{-\left(-10\right)±\sqrt{1700}}{2}$$

1700 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-10\right)±10\sqrt{17}}{2}$$

-10 ning teskarisi 10 ga teng.

$$x=\frac{10±10\sqrt{17}}{2}$$

x=\frac{10±10\sqrt{17}}{2} tenglamasini yeching, bunda ± musbat. 10 ni 10\sqrt{17} ga qo'shish.

$$x=\frac{10\sqrt{17}+10}{2}$$

10+10\sqrt{17} ni 2 ga bo'lish.

$$x=5\sqrt{17}+5$$

x=\frac{10±10\sqrt{17}}{2} tenglamasini yeching, bunda ± manfiy. 10 dan 10\sqrt{17} ni ayirish.

$$x=\frac{10-10\sqrt{17}}{2}$$

10-10\sqrt{17} ni 2 ga bo'lish.

$$x=5-5\sqrt{17}$$

Tenglama yechildi.

$$x=5\sqrt{17}+5 x=5-5\sqrt{17}$$

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.

$$x^{2}-10x-400=0$$

400 ni tenglamaning ikkala tarafiga qo'shish.

$$x^{2}-10x-400-\left(-400\right)=-\left(-400\right)$$

O‘zidan -400 ayirilsa 0 qoladi.

$$x^{2}-10x=-\left(-400\right)$$

0 dan -400 ni ayirish.

$$x^{2}-10x=400$$

-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.

$$x^{2}-10x+\left(-5\right)^{2}=400+\left(-5\right)^{2}$$

-5 kvadratini chiqarish.

$$x^{2}-10x+25=400+25$$

400 ni 25 ga qo'shish.

$$x^{2}-10x+25=425$$

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.

$$\left(x-5\right)^{2}=425$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-5\right)^{2}}=\sqrt{425}$$

Qisqartirish.

$$x-5=5\sqrt{17} x-5=-5\sqrt{17}$$

5 ni tenglamaning ikkala tarafiga qo'shish.

$$x=5\sqrt{17}+5 x=5-5\sqrt{17}$$