x^2-1/10x-3/10=0 eching

$${ x }^{ 2 } - \frac{ 1 }{ 10 } x- \frac{ 3 }{ 10 } =0$$

x uchun yechish

$$x=-\frac{1}{2}=-0,5$$

$$x=\frac{3}{5}=0,6$$

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.

$$x^{2}-\frac{1}{10}x-\frac{3}{10}=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 1 ni a, -\frac{1}{10} ni b va -\frac{3}{10} ni c bilan almashtiring.

$$x=\frac{-\left(-\frac{1}{10}\right)±\sqrt{\left(-\frac{1}{10}\right)^{2}-4\left(-\frac{3}{10}\right)}}{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{1}{10} kvadratini chiqarish.

$$x=\frac{-\left(-\frac{1}{10}\right)±\sqrt{\frac{1}{100}-4\left(-\frac{3}{10}\right)}}{2}$$

-4 ni -\frac{3}{10} marotabaga ko'paytirish.

$$x=\frac{-\left(-\frac{1}{10}\right)±\sqrt{\frac{1}{100}+\frac{6}{5}}}{2}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{1}{100} ni \frac{6}{5} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x=\frac{-\left(-\frac{1}{10}\right)±\sqrt{\frac{121}{100}}}{2}$$

\frac{121}{100} ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-\frac{1}{10}\right)±\frac{11}{10}}{2}$$

-\frac{1}{10} ning teskarisi \frac{1}{10} ga teng.

$$x=\frac{\frac{1}{10}±\frac{11}{10}}{2}$$

x=\frac{\frac{1}{10}±\frac{11}{10}}{2} tenglamasini yeching, bunda ± musbat. Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{1}{10} ni \frac{11}{10} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x=\frac{\frac{6}{5}}{2}$$

\frac{6}{5} ni 2 ga bo'lish.

$$x=\frac{3}{5}$$

x=\frac{\frac{1}{10}±\frac{11}{10}}{2} tenglamasini yeching, bunda ± manfiy. Umumiy maxrajni topib va suratlarni ayirib \frac{11}{10} ni \frac{1}{10} dan ayirish. So'ngra imkoni boricha kasrni eng kichik shartga qisqartirish.

$$x=-\frac{1}{2}$$

Tenglama yechildi.

$$x=\frac{3}{5} x=-\frac{1}{2}$$

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.

$$x^{2}-\frac{1}{10}x-\frac{3}{10}=0$$

\frac{3}{10} ni tenglamaning ikkala tarafiga qo'shish.

$$x^{2}-\frac{1}{10}x-\frac{3}{10}-\left(-\frac{3}{10}\right)=-\left(-\frac{3}{10}\right)$$

O‘zidan -\frac{3}{10} ayirilsa 0 qoladi.

$$x^{2}-\frac{1}{10}x=-\left(-\frac{3}{10}\right)$$

0 dan -\frac{3}{10} ni ayirish.

$$x^{2}-\frac{1}{10}x=\frac{3}{10}$$

-\frac{1}{10} ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{1}{20} olish uchun. Keyin, -\frac{1}{20} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}-\frac{1}{10}x+\left(-\frac{1}{20}\right)^{2}=\frac{3}{10}+\left(-\frac{1}{20}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{1}{20} kvadratini chiqarish.

$$x^{2}-\frac{1}{10}x+\frac{1}{400}=\frac{3}{10}+\frac{1}{400}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{3}{10} ni \frac{1}{400} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}-\frac{1}{10}x+\frac{1}{400}=\frac{121}{400}$$

x^{2}-\frac{1}{10}x+\frac{1}{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.

$$\left(x-\frac{1}{20}\right)^{2}=\frac{121}{400}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{1}{20}\right)^{2}}=\sqrt{\frac{121}{400}}$$

Qisqartirish.

$$x-\frac{1}{20}=\frac{11}{20} x-\frac{1}{20}=-\frac{11}{20}$$

\frac{1}{20} ni tenglamaning ikkala tarafiga qo'shish.

$$x=\frac{3}{5} x=-\frac{1}{2}$$