300x^2+800x-800=0 eching

$$300 x ^ { 2 } + 800 x - 800 = 0$$

x uchun yechish

$$x=\frac{2\sqrt{10}-4}{3}\approx 0,774851773$$

$$x=\frac{-2\sqrt{10}-4}{3}\approx -3,44151844$$

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.

$$300x^{2}+800x-800=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 300 ni a, 800 ni b va -800 ni c bilan almashtiring.

$$x=\frac{-800±\sqrt{800^{2}-4\times 300\left(-800\right)}}{2\times 300}$$

800 kvadratini chiqarish.

$$x=\frac{-800±\sqrt{640000-4\times 300\left(-800\right)}}{2\times 300}$$

-4 ni 300 marotabaga ko'paytirish.

$$x=\frac{-800±\sqrt{640000-1200\left(-800\right)}}{2\times 300}$$

-1200 ni -800 marotabaga ko'paytirish.

$$x=\frac{-800±\sqrt{640000+960000}}{2\times 300}$$

640000 ni 960000 ga qo'shish.

$$x=\frac{-800±\sqrt{1600000}}{2\times 300}$$

1600000 ning kvadrat ildizini chiqarish.

$$x=\frac{-800±400\sqrt{10}}{2\times 300}$$

2 ni 300 marotabaga ko'paytirish.

$$x=\frac{-800±400\sqrt{10}}{600}$$

x=\frac{-800±400\sqrt{10}}{600} tenglamasini yeching, bunda ± musbat. -800 ni 400\sqrt{10} ga qo'shish.

$$x=\frac{400\sqrt{10}-800}{600}$$

-800+400\sqrt{10} ni 600 ga bo'lish.

$$x=\frac{2\sqrt{10}-4}{3}$$

x=\frac{-800±400\sqrt{10}}{600} tenglamasini yeching, bunda ± manfiy. -800 dan 400\sqrt{10} ni ayirish.

$$x=\frac{-400\sqrt{10}-800}{600}$$

-800-400\sqrt{10} ni 600 ga bo'lish.

$$x=\frac{-2\sqrt{10}-4}{3}$$

Tenglama yechildi.

$$x=\frac{2\sqrt{10}-4}{3} x=\frac{-2\sqrt{10}-4}{3}$$

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.

$$300x^{2}+800x-800=0$$

800 ni tenglamaning ikkala tarafiga qo'shish.

$$300x^{2}+800x-800-\left(-800\right)=-\left(-800\right)$$

O‘zidan -800 ayirilsa 0 qoladi.

$$300x^{2}+800x=-\left(-800\right)$$

0 dan -800 ni ayirish.

$$300x^{2}+800x=800$$

Ikki tarafini 300 ga bo‘ling.

$$\frac{300x^{2}+800x}{300}=\frac{800}{300}$$

300 ga bo'lish 300 ga ko'paytirishni bekor qiladi.

$$x^{2}+\frac{800}{300}x=\frac{800}{300}$$

\frac{800}{300} ulushini 100 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x^{2}+\frac{8}{3}x=\frac{800}{300}$$

\frac{800}{300} ulushini 100 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x^{2}+\frac{8}{3}x=\frac{8}{3}$$

\frac{8}{3} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{4}{3} olish uchun. Keyin, \frac{4}{3} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+\frac{8}{3}x+\left(\frac{4}{3}\right)^{2}=\frac{8}{3}+\left(\frac{4}{3}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{4}{3} kvadratini chiqarish.

$$x^{2}+\frac{8}{3}x+\frac{16}{9}=\frac{8}{3}+\frac{16}{9}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{8}{3} ni \frac{16}{9} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}+\frac{8}{3}x+\frac{16}{9}=\frac{40}{9}$$

x^{2}+\frac{8}{3}x+\frac{16}{9} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+\frac{4}{3}\right)^{2}=\frac{40}{9}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{4}{3}\right)^{2}}=\sqrt{\frac{40}{9}}$$

Qisqartirish.

$$x+\frac{4}{3}=\frac{2\sqrt{10}}{3} x+\frac{4}{3}=-\frac{2\sqrt{10}}{3}$$

Tenglamaning ikkala tarafidan \frac{4}{3} ni ayirish.

$$x=\frac{2\sqrt{10}-4}{3} x=\frac{-2\sqrt{10}-4}{3}$$