(x+3)^2+5x=8 eching

$$( x + 3 ) ^ { 2 } + 5 x = 8$$

x uchun yechish

$$x=\frac{3\sqrt{13}-11}{2}\approx -0,091673087$$

$$x=\frac{-3\sqrt{13}-11}{2}\approx -10,908326913$$

Kvadrat tenglamalaridan foydalanish qoidalari

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

$$x^{2}+6x+9+5x=8$$

11x ni olish uchun 6x va 5x ni birlashtirish.

$$x^{2}+11x+9=8$$

Ikkala tarafdan 8 ni ayirish.

$$x^{2}+11x+9-8=0$$

1 olish uchun 9 dan 8 ni ayirish.

$$x^{2}+11x+1=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 1 ni a, 11 ni b va 1 ni c bilan almashtiring.

$$x=\frac{-11±\sqrt{11^{2}-4}}{2}$$

11 kvadratini chiqarish.

$$x=\frac{-11±\sqrt{121-4}}{2}$$

121 ni -4 ga qo'shish.

$$x=\frac{-11±\sqrt{117}}{2}$$

117 ning kvadrat ildizini chiqarish.

$$x=\frac{-11±3\sqrt{13}}{2}$$

x=\frac{-11±3\sqrt{13}}{2} tenglamasini yeching, bunda ± musbat. -11 ni 3\sqrt{13} ga qo'shish.

$$x=\frac{3\sqrt{13}-11}{2}$$

x=\frac{-11±3\sqrt{13}}{2} tenglamasini yeching, bunda ± manfiy. -11 dan 3\sqrt{13} ni ayirish.

$$x=\frac{-3\sqrt{13}-11}{2}$$

Tenglama yechildi.

$$x=\frac{3\sqrt{13}-11}{2} x=\frac{-3\sqrt{13}-11}{2}$$

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.

$$x^{2}+6x+9+5x=8$$

11x ni olish uchun 6x va 5x ni birlashtirish.

$$x^{2}+11x+9=8$$

Ikkala tarafdan 9 ni ayirish.

$$x^{2}+11x=8-9$$

-1 olish uchun 8 dan 9 ni ayirish.

$$x^{2}+11x=-1$$

11 ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{11}{2} olish uchun. Keyin, \frac{11}{2} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+11x+\left(\frac{11}{2}\right)^{2}=-1+\left(\frac{11}{2}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{11}{2} kvadratini chiqarish.

$$x^{2}+11x+\frac{121}{4}=-1+\frac{121}{4}$$

-1 ni \frac{121}{4} ga qo'shish.

$$x^{2}+11x+\frac{121}{4}=\frac{117}{4}$$

x^{2}+11x+\frac{121}{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.

$$\left(x+\frac{11}{2}\right)^{2}=\frac{117}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{11}{2}\right)^{2}}=\sqrt{\frac{117}{4}}$$

Qisqartirish.

$$x+\frac{11}{2}=\frac{3\sqrt{13}}{2} x+\frac{11}{2}=-\frac{3\sqrt{13}}{2}$$

Tenglamaning ikkala tarafidan \frac{11}{2} ni ayirish.

$$x=\frac{3\sqrt{13}-11}{2} x=\frac{-3\sqrt{13}-11}{2}$$