2x^2=-35x-1 eching

$$2 x ^ { 2 } = - 35 x - 1$$

x uchun yechish

$$x=\frac{\sqrt{1217}-35}{4}\approx -0,028618229$$

$$x=\frac{-\sqrt{1217}-35}{4}\approx -17,471381771$$

Kvadrat tenglamalaridan foydalanish qoidalari

35x ni ikki tarafga qo’shing.

$$2x^{2}+35x=-1$$

1 ni ikki tarafga qo’shing.

$$2x^{2}+35x+1=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 2 ni a, 35 ni b va 1 ni c bilan almashtiring.

$$x=\frac{-35±\sqrt{35^{2}-4\times 2}}{2\times 2}$$

35 kvadratini chiqarish.

$$x=\frac{-35±\sqrt{1225-4\times 2}}{2\times 2}$$

-4 ni 2 marotabaga ko'paytirish.

$$x=\frac{-35±\sqrt{1225-8}}{2\times 2}$$

1225 ni -8 ga qo'shish.

$$x=\frac{-35±\sqrt{1217}}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$x=\frac{-35±\sqrt{1217}}{4}$$

x=\frac{-35±\sqrt{1217}}{4} tenglamasini yeching, bunda ± musbat. -35 ni \sqrt{1217} ga qo'shish.

$$x=\frac{\sqrt{1217}-35}{4}$$

x=\frac{-35±\sqrt{1217}}{4} tenglamasini yeching, bunda ± manfiy. -35 dan \sqrt{1217} ni ayirish.

$$x=\frac{-\sqrt{1217}-35}{4}$$

Tenglama yechildi.

$$x=\frac{\sqrt{1217}-35}{4} x=\frac{-\sqrt{1217}-35}{4}$$

Kvadratni yakunlash bo'yicha qadamlar

35x ni ikki tarafga qo’shing.

$$2x^{2}+35x=-1$$

Ikki tarafini 2 ga bo‘ling.

$$\frac{2x^{2}+35x}{2}=-\frac{1}{2}$$

2 ga bo'lish 2 ga ko'paytirishni bekor qiladi.

$$x^{2}+\frac{35}{2}x=-\frac{1}{2}$$

\frac{35}{2} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{35}{4} olish uchun. Keyin, \frac{35}{4} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+\frac{35}{2}x+\left(\frac{35}{4}\right)^{2}=-\frac{1}{2}+\left(\frac{35}{4}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{35}{4} kvadratini chiqarish.

$$x^{2}+\frac{35}{2}x+\frac{1225}{16}=-\frac{1}{2}+\frac{1225}{16}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali -\frac{1}{2} ni \frac{1225}{16} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}+\frac{35}{2}x+\frac{1225}{16}=\frac{1217}{16}$$

x^{2}+\frac{35}{2}x+\frac{1225}{16} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+\frac{35}{4}\right)^{2}=\frac{1217}{16}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{35}{4}\right)^{2}}=\sqrt{\frac{1217}{16}}$$

Qisqartirish.

$$x+\frac{35}{4}=\frac{\sqrt{1217}}{4} x+\frac{35}{4}=-\frac{\sqrt{1217}}{4}$$

Tenglamaning ikkala tarafidan \frac{35}{4} ni ayirish.

$$x=\frac{\sqrt{1217}-35}{4} x=\frac{-\sqrt{1217}-35}{4}$$