10x^2+3x-3=0 eching

$$10 x ^ { 2 } + 3 x - 3 = 0$$

x uchun yechish

$$x=\frac{\sqrt{129}-3}{20}\approx 0,417890835$$

$$x=\frac{-\sqrt{129}-3}{20}\approx -0,717890835$$

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.

$$10x^{2}+3x-3=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 10 ni a, 3 ni b va -3 ni c bilan almashtiring.

$$x=\frac{-3±\sqrt{3^{2}-4\times 10\left(-3\right)}}{2\times 10}$$

3 kvadratini chiqarish.

$$x=\frac{-3±\sqrt{9-4\times 10\left(-3\right)}}{2\times 10}$$

-4 ni 10 marotabaga ko'paytirish.

$$x=\frac{-3±\sqrt{9-40\left(-3\right)}}{2\times 10}$$

-40 ni -3 marotabaga ko'paytirish.

$$x=\frac{-3±\sqrt{9+120}}{2\times 10}$$

9 ni 120 ga qo'shish.

$$x=\frac{-3±\sqrt{129}}{2\times 10}$$

2 ni 10 marotabaga ko'paytirish.

$$x=\frac{-3±\sqrt{129}}{20}$$

x=\frac{-3±\sqrt{129}}{20} tenglamasini yeching, bunda ± musbat. -3 ni \sqrt{129} ga qo'shish.

$$x=\frac{\sqrt{129}-3}{20}$$

x=\frac{-3±\sqrt{129}}{20} tenglamasini yeching, bunda ± manfiy. -3 dan \sqrt{129} ni ayirish.

$$x=\frac{-\sqrt{129}-3}{20}$$

Tenglama yechildi.

$$x=\frac{\sqrt{129}-3}{20} x=\frac{-\sqrt{129}-3}{20}$$

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.

$$10x^{2}+3x-3=0$$

3 ni tenglamaning ikkala tarafiga qo'shish.

$$10x^{2}+3x-3-\left(-3\right)=-\left(-3\right)$$

O‘zidan -3 ayirilsa 0 qoladi.

$$10x^{2}+3x=-\left(-3\right)$$

0 dan -3 ni ayirish.

$$10x^{2}+3x=3$$

Ikki tarafini 10 ga bo‘ling.

$$\frac{10x^{2}+3x}{10}=\frac{3}{10}$$

10 ga bo'lish 10 ga ko'paytirishni bekor qiladi.

$$x^{2}+\frac{3}{10}x=\frac{3}{10}$$

\frac{3}{10} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{3}{20} olish uchun. Keyin, \frac{3}{20} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+\frac{3}{10}x+\left(\frac{3}{20}\right)^{2}=\frac{3}{10}+\left(\frac{3}{20}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{3}{20} kvadratini chiqarish.

$$x^{2}+\frac{3}{10}x+\frac{9}{400}=\frac{3}{10}+\frac{9}{400}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{3}{10} ni \frac{9}{400} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}+\frac{3}{10}x+\frac{9}{400}=\frac{129}{400}$$

x^{2}+\frac{3}{10}x+\frac{9}{400} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+\frac{3}{20}\right)^{2}=\frac{129}{400}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{3}{20}\right)^{2}}=\sqrt{\frac{129}{400}}$$

Qisqartirish.

$$x+\frac{3}{20}=\frac{\sqrt{129}}{20} x+\frac{3}{20}=-\frac{\sqrt{129}}{20}$$

Tenglamaning ikkala tarafidan \frac{3}{20} ni ayirish.

$$x=\frac{\sqrt{129}-3}{20} x=\frac{-\sqrt{129}-3}{20}$$