2+1=left(2x+3right)left(5x-3right) eching

$$2+1 = \left( 2x+3 \right) \left( 5x-3 \right)$$

x uchun yechish

$$x=\frac{\sqrt{561}-9}{20}\approx 0,734271928$$

$$x=\frac{-\sqrt{561}-9}{20}\approx -1,634271928$$

Kvadrat tenglamalaridan foydalanish qoidalari

3 olish uchun 2 va 1'ni qo'shing.

$$3=\left(2x+3\right)\left(5x-3\right)$$

2x+3 ga 5x-3 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

$$3=10x^{2}+9x-9$$

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$10x^{2}+9x-9=3$$

Ikkala tarafdan 3 ni ayirish.

$$10x^{2}+9x-9-3=0$$

-12 olish uchun -9 dan 3 ni ayirish.

$$10x^{2}+9x-12=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 10 ni a, 9 ni b va -12 ni c bilan almashtiring.

$$x=\frac{-9±\sqrt{9^{2}-4\times 10\left(-12\right)}}{2\times 10}$$

9 kvadratini chiqarish.

$$x=\frac{-9±\sqrt{81-4\times 10\left(-12\right)}}{2\times 10}$$

-4 ni 10 marotabaga ko'paytirish.

$$x=\frac{-9±\sqrt{81-40\left(-12\right)}}{2\times 10}$$

-40 ni -12 marotabaga ko'paytirish.

$$x=\frac{-9±\sqrt{81+480}}{2\times 10}$$

81 ni 480 ga qo'shish.

$$x=\frac{-9±\sqrt{561}}{2\times 10}$$

2 ni 10 marotabaga ko'paytirish.

$$x=\frac{-9±\sqrt{561}}{20}$$

x=\frac{-9±\sqrt{561}}{20} tenglamasini yeching, bunda ± musbat. -9 ni \sqrt{561} ga qo'shish.

$$x=\frac{\sqrt{561}-9}{20}$$

x=\frac{-9±\sqrt{561}}{20} tenglamasini yeching, bunda ± manfiy. -9 dan \sqrt{561} ni ayirish.

$$x=\frac{-\sqrt{561}-9}{20}$$

Tenglama yechildi.

$$x=\frac{\sqrt{561}-9}{20} x=\frac{-\sqrt{561}-9}{20}$$

Kvadratni yakunlash bo'yicha qadamlar

3 olish uchun 2 va 1'ni qo'shing.

$$3=\left(2x+3\right)\left(5x-3\right)$$

2x+3 ga 5x-3 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

$$3=10x^{2}+9x-9$$

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$10x^{2}+9x-9=3$$

9 ni ikki tarafga qo’shing.

$$10x^{2}+9x=3+9$$

12 olish uchun 3 va 9'ni qo'shing.

$$10x^{2}+9x=12$$

Ikki tarafini 10 ga bo‘ling.

$$\frac{10x^{2}+9x}{10}=\frac{12}{10}$$

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

$$x^{2}+\frac{9}{10}x=\frac{12}{10}$$

\frac{12}{10} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x^{2}+\frac{9}{10}x=\frac{6}{5}$$

\frac{9}{10} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{9}{20} olish uchun. Keyin, \frac{9}{20} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+\frac{9}{10}x+\left(\frac{9}{20}\right)^{2}=\frac{6}{5}+\left(\frac{9}{20}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{9}{20} kvadratini chiqarish.

$$x^{2}+\frac{9}{10}x+\frac{81}{400}=\frac{6}{5}+\frac{81}{400}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{6}{5} ni \frac{81}{400} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}+\frac{9}{10}x+\frac{81}{400}=\frac{561}{400}$$

x^{2}+\frac{9}{10}x+\frac{81}{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.

$$\left(x+\frac{9}{20}\right)^{2}=\frac{561}{400}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{9}{20}\right)^{2}}=\sqrt{\frac{561}{400}}$$

Qisqartirish.

$$x+\frac{9}{20}=\frac{\sqrt{561}}{20} x+\frac{9}{20}=-\frac{\sqrt{561}}{20}$$

Tenglamaning ikkala tarafidan \frac{9}{20} ni ayirish.

$$x=\frac{\sqrt{561}-9}{20} x=\frac{-\sqrt{561}-9}{20}$$