left(2x-3right)^2-left(x-5right)^2=-23 eching

$${ \left(2x-3 \right) }^{ 2 } - { \left(x-5 \right) }^{ 2 } =-23$$

x uchun yechish (complex solution)

$$x=\frac{1+2\sqrt{5}i}{3}\approx 0,333333333+1,490711985i$$

$$x=\frac{-2\sqrt{5}i+1}{3}\approx 0,333333333-1,490711985i$$

Kvadrat tenglamalaridan foydalanish qoidalari

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

$$4x^{2}-12x+9-\left(x-5\right)^{2}=-23$$

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(x-5\right)^{2} kengaytirilishi uchun ishlating.

$$4x^{2}-12x+9-\left(x^{2}-10x+25\right)=-23$$

x^{2}-10x+25 teskarisini topish uchun har birining teskarisini toping.

$$4x^{2}-12x+9-x^{2}+10x-25=-23$$

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

$$3x^{2}-12x+9+10x-25=-23$$

-2x ni olish uchun -12x va 10x ni birlashtirish.

$$3x^{2}-2x+9-25=-23$$

-16 olish uchun 9 dan 25 ni ayirish.

$$3x^{2}-2x-16=-23$$

23 ni ikki tarafga qo’shing.

$$3x^{2}-2x-16+23=0$$

7 olish uchun -16 va 23'ni qo'shing.

$$3x^{2}-2x+7=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 3 ni a, -2 ni b va 7 ni c bilan almashtiring.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 3\times 7}}{2\times 3}$$

-2 kvadratini chiqarish.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\times 3\times 7}}{2\times 3}$$

-4 ni 3 marotabaga ko'paytirish.

$$x=\frac{-\left(-2\right)±\sqrt{4-12\times 7}}{2\times 3}$$

-12 ni 7 marotabaga ko'paytirish.

$$x=\frac{-\left(-2\right)±\sqrt{4-84}}{2\times 3}$$

4 ni -84 ga qo'shish.

$$x=\frac{-\left(-2\right)±\sqrt{-80}}{2\times 3}$$

-80 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-2\right)±4\sqrt{5}i}{2\times 3}$$

-2 ning teskarisi 2 ga teng.

$$x=\frac{2±4\sqrt{5}i}{2\times 3}$$

2 ni 3 marotabaga ko'paytirish.

$$x=\frac{2±4\sqrt{5}i}{6}$$

x=\frac{2±4\sqrt{5}i}{6} tenglamasini yeching, bunda ± musbat. 2 ni 4i\sqrt{5} ga qo'shish.

$$x=\frac{2+4\sqrt{5}i}{6}$$

2+4i\sqrt{5} ni 6 ga bo'lish.

$$x=\frac{1+2\sqrt{5}i}{3}$$

x=\frac{2±4\sqrt{5}i}{6} tenglamasini yeching, bunda ± manfiy. 2 dan 4i\sqrt{5} ni ayirish.

$$x=\frac{-4\sqrt{5}i+2}{6}$$

2-4i\sqrt{5} ni 6 ga bo'lish.

$$x=\frac{-2\sqrt{5}i+1}{3}$$

Tenglama yechildi.

$$x=\frac{1+2\sqrt{5}i}{3} x=\frac{-2\sqrt{5}i+1}{3}$$

Kvadratni yakunlash bo'yicha qadamlar

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

$$4x^{2}-12x+9-\left(x-5\right)^{2}=-23$$

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(x-5\right)^{2} kengaytirilishi uchun ishlating.

$$4x^{2}-12x+9-\left(x^{2}-10x+25\right)=-23$$

x^{2}-10x+25 teskarisini topish uchun har birining teskarisini toping.

$$4x^{2}-12x+9-x^{2}+10x-25=-23$$

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

$$3x^{2}-12x+9+10x-25=-23$$

-2x ni olish uchun -12x va 10x ni birlashtirish.

$$3x^{2}-2x+9-25=-23$$

-16 olish uchun 9 dan 25 ni ayirish.

$$3x^{2}-2x-16=-23$$

16 ni ikki tarafga qo’shing.

$$3x^{2}-2x=-23+16$$

-7 olish uchun -23 va 16'ni qo'shing.

$$3x^{2}-2x=-7$$

Ikki tarafini 3 ga bo‘ling.

$$\frac{3x^{2}-2x}{3}=-\frac{7}{3}$$

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

$$x^{2}-\frac{2}{3}x=-\frac{7}{3}$$

-\frac{2}{3} ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{1}{3} olish uchun. Keyin, -\frac{1}{3} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}-\frac{2}{3}x+\left(-\frac{1}{3}\right)^{2}=-\frac{7}{3}+\left(-\frac{1}{3}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{1}{3} kvadratini chiqarish.

$$x^{2}-\frac{2}{3}x+\frac{1}{9}=-\frac{7}{3}+\frac{1}{9}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali -\frac{7}{3} ni \frac{1}{9} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}-\frac{2}{3}x+\frac{1}{9}=-\frac{20}{9}$$

x^{2}-\frac{2}{3}x+\frac{1}{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.

$$\left(x-\frac{1}{3}\right)^{2}=-\frac{20}{9}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{1}{3}\right)^{2}}=\sqrt{-\frac{20}{9}}$$

Qisqartirish.

$$x-\frac{1}{3}=\frac{2\sqrt{5}i}{3} x-\frac{1}{3}=-\frac{2\sqrt{5}i}{3}$$

\frac{1}{3} ni tenglamaning ikkala tarafiga qo'shish.

$$x=\frac{1+2\sqrt{5}i}{3} x=\frac{-2\sqrt{5}i+1}{3}$$