x^2+x-2x-(2x^2-5)=0 eching

$${ x }^{ 2 } +x-2x-(2 { x }^{ 2 } -5)=0$$

x uchun yechish

$$x = \frac{\sqrt{21} - 1}{2} \approx 1,791287847$$

$$x=\frac{-\sqrt{21}-1}{2}\approx -2,791287847$$

Kvadrat tenglamalaridan foydalanish qoidalari

-x ni olish uchun x va -2x ni birlashtirish.

$$x^{2}-x-\left(2x^{2}-5\right)=0$$

2x^{2}-5 teskarisini topish uchun har birining teskarisini toping.

$$x^{2}-x-2x^{2}+5=0$$

-x^{2} ni olish uchun x^{2} va -2x^{2} ni birlashtirish.

$$-x^{2}-x+5=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} -1 ni a, -1 ni b va 5 ni c bilan almashtiring.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\left(-1\right)\times 5}}{2\left(-1\right)}$$

-4 ni -1 marotabaga ko'paytirish.

$$x=\frac{-\left(-1\right)±\sqrt{1+4\times 5}}{2\left(-1\right)}$$

4 ni 5 marotabaga ko'paytirish.

$$x=\frac{-\left(-1\right)±\sqrt{1+20}}{2\left(-1\right)}$$

1 ni 20 ga qo'shish.

$$x=\frac{-\left(-1\right)±\sqrt{21}}{2\left(-1\right)}$$

-1 ning teskarisi 1 ga teng.

$$x=\frac{1±\sqrt{21}}{2\left(-1\right)}$$

2 ni -1 marotabaga ko'paytirish.

$$x=\frac{1±\sqrt{21}}{-2}$$

x=\frac{1±\sqrt{21}}{-2} tenglamasini yeching, bunda ± musbat. 1 ni \sqrt{21} ga qo'shish.

$$x=\frac{\sqrt{21}+1}{-2}$$

1+\sqrt{21} ni -2 ga bo'lish.

$$x=\frac{-\sqrt{21}-1}{2}$$

x=\frac{1±\sqrt{21}}{-2} tenglamasini yeching, bunda ± manfiy. 1 dan \sqrt{21} ni ayirish.

$$x=\frac{1-\sqrt{21}}{-2}$$

1-\sqrt{21} ni -2 ga bo'lish.

$$x=\frac{\sqrt{21}-1}{2}$$

Tenglama yechildi.

$$x=\frac{-\sqrt{21}-1}{2} x=\frac{\sqrt{21}-1}{2}$$

Kvadratni yakunlash bo'yicha qadamlar

-x ni olish uchun x va -2x ni birlashtirish.

$$x^{2}-x-\left(2x^{2}-5\right)=0$$

2x^{2}-5 teskarisini topish uchun har birining teskarisini toping.

$$x^{2}-x-2x^{2}+5=0$$

-x^{2} ni olish uchun x^{2} va -2x^{2} ni birlashtirish.

$$-x^{2}-x+5=0$$

Ikkala tarafdan 5 ni ayirish. Har qanday sonni noldan ayirsangiz, o‘zining manfiyi chiqadi.

$$-x^{2}-x=-5$$

Ikki tarafini -1 ga bo‘ling.

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

-1 ga bo'lish -1 ga ko'paytirishni bekor qiladi.

$$x^{2}+\left(-\frac{1}{-1}\right)x=-\frac{5}{-1}$$

-1 ni -1 ga bo'lish.

$$x^{2}+x=-\frac{5}{-1}$$

-5 ni -1 ga bo'lish.

$$x^{2}+x=5$$

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

$$x^{2}+x+\left(\frac{1}{2}\right)^{2}=5+\left(\frac{1}{2}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{1}{2} kvadratini chiqarish.

$$x^{2}+x+\frac{1}{4}=5+\frac{1}{4}$$

5 ni \frac{1}{4} ga qo'shish.

$$x^{2}+x+\frac{1}{4}=\frac{21}{4}$$

x^{2}+x+\frac{1}{4} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+\frac{1}{2}\right)^{2}=\frac{21}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{21}{4}}$$

Qisqartirish.

$$x+\frac{1}{2}=\frac{\sqrt{21}}{2} x+\frac{1}{2}=-\frac{\sqrt{21}}{2}$$

Tenglamaning ikkala tarafidan \frac{1}{2} ni ayirish.

$$x=\frac{\sqrt{21}-1}{2} x=\frac{-\sqrt{21}-1}{2}$$