(x-3)^2-(2x+1)^2=5 eching

$$( x - 3 ) ^ { 2 } - ( 2 x + 1 ) ^ { 2 } = 5$$

x uchun yechish

$$x=\frac{\sqrt{34}-5}{3}\approx 0,276983965$$

$$x=\frac{-\sqrt{34}-5}{3}\approx -3,610317298$$

Kvadrat tenglamalaridan foydalanish qoidalari

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(x-3\right)^{2} kengaytirilishi uchun ishlating.

$$x^{2}-6x+9-\left(2x+1\right)^{2}=5$$

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(2x+1\right)^{2} kengaytirilishi uchun ishlating.

$$x^{2}-6x+9-\left(4x^{2}+4x+1\right)=5$$

4x^{2}+4x+1 teskarisini topish uchun har birining teskarisini toping.

$$x^{2}-6x+9-4x^{2}-4x-1=5$$

-3x^{2} ni olish uchun x^{2} va -4x^{2} ni birlashtirish.

$$-3x^{2}-6x+9-4x-1=5$$

-10x ni olish uchun -6x va -4x ni birlashtirish.

$$-3x^{2}-10x+9-1=5$$

8 olish uchun 9 dan 1 ni ayirish.

$$-3x^{2}-10x+8=5$$

Ikkala tarafdan 5 ni ayirish.

$$-3x^{2}-10x+8-5=0$$

3 olish uchun 8 dan 5 ni ayirish.

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

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

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

-10 kvadratini chiqarish.

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

-4 ni -3 marotabaga ko'paytirish.

$$x=\frac{-\left(-10\right)±\sqrt{100+12\times 3}}{2\left(-3\right)}$$

12 ni 3 marotabaga ko'paytirish.

$$x=\frac{-\left(-10\right)±\sqrt{100+36}}{2\left(-3\right)}$$

100 ni 36 ga qo'shish.

$$x=\frac{-\left(-10\right)±\sqrt{136}}{2\left(-3\right)}$$

136 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-10\right)±2\sqrt{34}}{2\left(-3\right)}$$

-10 ning teskarisi 10 ga teng.

$$x=\frac{10±2\sqrt{34}}{2\left(-3\right)}$$

2 ni -3 marotabaga ko'paytirish.

$$x=\frac{10±2\sqrt{34}}{-6}$$

x=\frac{10±2\sqrt{34}}{-6} tenglamasini yeching, bunda ± musbat. 10 ni 2\sqrt{34} ga qo'shish.

$$x=\frac{2\sqrt{34}+10}{-6}$$

10+2\sqrt{34} ni -6 ga bo'lish.

$$x=\frac{-\sqrt{34}-5}{3}$$

x=\frac{10±2\sqrt{34}}{-6} tenglamasini yeching, bunda ± manfiy. 10 dan 2\sqrt{34} ni ayirish.

$$x=\frac{10-2\sqrt{34}}{-6}$$

10-2\sqrt{34} ni -6 ga bo'lish.

$$x=\frac{\sqrt{34}-5}{3}$$

Tenglama yechildi.

$$x=\frac{-\sqrt{34}-5}{3} x=\frac{\sqrt{34}-5}{3}$$

Kvadratni yakunlash bo'yicha qadamlar

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(x-3\right)^{2} kengaytirilishi uchun ishlating.

$$x^{2}-6x+9-\left(2x+1\right)^{2}=5$$

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(2x+1\right)^{2} kengaytirilishi uchun ishlating.

$$x^{2}-6x+9-\left(4x^{2}+4x+1\right)=5$$

4x^{2}+4x+1 teskarisini topish uchun har birining teskarisini toping.

$$x^{2}-6x+9-4x^{2}-4x-1=5$$

-3x^{2} ni olish uchun x^{2} va -4x^{2} ni birlashtirish.

$$-3x^{2}-6x+9-4x-1=5$$

-10x ni olish uchun -6x va -4x ni birlashtirish.

$$-3x^{2}-10x+9-1=5$$

8 olish uchun 9 dan 1 ni ayirish.

$$-3x^{2}-10x+8=5$$

Ikkala tarafdan 8 ni ayirish.

$$-3x^{2}-10x=5-8$$

-3 olish uchun 5 dan 8 ni ayirish.

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

Ikki tarafini -3 ga bo‘ling.

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

-3 ga bo'lish -3 ga ko'paytirishni bekor qiladi.

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

-10 ni -3 ga bo'lish.

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

-3 ni -3 ga bo'lish.

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

\frac{10}{3} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{5}{3} olish uchun. Keyin, \frac{5}{3} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+\frac{10}{3}x+\left(\frac{5}{3}\right)^{2}=1+\left(\frac{5}{3}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{5}{3} kvadratini chiqarish.

$$x^{2}+\frac{10}{3}x+\frac{25}{9}=1+\frac{25}{9}$$

1 ni \frac{25}{9} ga qo'shish.

$$x^{2}+\frac{10}{3}x+\frac{25}{9}=\frac{34}{9}$$

x^{2}+\frac{10}{3}x+\frac{25}{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.

$$\left(x+\frac{5}{3}\right)^{2}=\frac{34}{9}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{5}{3}\right)^{2}}=\sqrt{\frac{34}{9}}$$

Qisqartirish.

$$x+\frac{5}{3}=\frac{\sqrt{34}}{3} x+\frac{5}{3}=-\frac{\sqrt{34}}{3}$$

Tenglamaning ikkala tarafidan \frac{5}{3} ni ayirish.

$$x=\frac{\sqrt{34}-5}{3} x=\frac{-\sqrt{34}-5}{3}$$