x^2+20x+17=-3 eching

$${ x }^{ 2 } +20x+17=-3$$

x uchun yechish

$$x=4\sqrt{5}-10\approx -1,05572809$$

$$x=-4\sqrt{5}-10\approx -18,94427191$$

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.

$$x^{2}+20x+17=-3$$

3 ni tenglamaning ikkala tarafiga qo'shish.

$$x^{2}+20x+17-\left(-3\right)=-3-\left(-3\right)$$

O‘zidan -3 ayirilsa 0 qoladi.

$$x^{2}+20x+17-\left(-3\right)=0$$

17 dan -3 ni ayirish.

$$x^{2}+20x+20=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 1 ni a, 20 ni b va 20 ni c bilan almashtiring.

$$x=\frac{-20±\sqrt{20^{2}-4\times 20}}{2}$$

20 kvadratini chiqarish.

$$x=\frac{-20±\sqrt{400-4\times 20}}{2}$$

-4 ni 20 marotabaga ko'paytirish.

$$x=\frac{-20±\sqrt{400-80}}{2}$$

400 ni -80 ga qo'shish.

$$x=\frac{-20±\sqrt{320}}{2}$$

320 ning kvadrat ildizini chiqarish.

$$x=\frac{-20±8\sqrt{5}}{2}$$

x=\frac{-20±8\sqrt{5}}{2} tenglamasini yeching, bunda ± musbat. -20 ni 8\sqrt{5} ga qo'shish.

$$x=\frac{8\sqrt{5}-20}{2}$$

-20+8\sqrt{5} ni 2 ga bo'lish.

$$x=4\sqrt{5}-10$$

x=\frac{-20±8\sqrt{5}}{2} tenglamasini yeching, bunda ± manfiy. -20 dan 8\sqrt{5} ni ayirish.

$$x=\frac{-8\sqrt{5}-20}{2}$$

-20-8\sqrt{5} ni 2 ga bo'lish.

$$x=-4\sqrt{5}-10$$

Tenglama yechildi.

$$x=4\sqrt{5}-10 x=-4\sqrt{5}-10$$

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.

$$x^{2}+20x+17=-3$$

Tenglamaning ikkala tarafidan 17 ni ayirish.

$$x^{2}+20x+17-17=-3-17$$

O‘zidan 17 ayirilsa 0 qoladi.

$$x^{2}+20x=-3-17$$

-3 dan 17 ni ayirish.

$$x^{2}+20x=-20$$

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

$$x^{2}+20x+10^{2}=-20+10^{2}$$

10 kvadratini chiqarish.

$$x^{2}+20x+100=-20+100$$

-20 ni 100 ga qo'shish.

$$x^{2}+20x+100=80$$

x^{2}+20x+100 omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+10\right)^{2}=80$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+10\right)^{2}}=\sqrt{80}$$

Qisqartirish.

$$x+10=4\sqrt{5} x+10=-4\sqrt{5}$$

Tenglamaning ikkala tarafidan 10 ni ayirish.

$$x=4\sqrt{5}-10 x=-4\sqrt{5}-10$$