-5z^2-3z-11=-6z^2 eching

$$- 5 z ^ { 2 } - 3 z - 11 = - 6 z ^ { 2 }$$

z uchun yechish

$$z = \frac{\sqrt{53} + 3}{2} \approx 5,140054945$$

$$z=\frac{3-\sqrt{53}}{2}\approx -2,140054945$$

Kvadrat tenglamalaridan foydalanish qoidalari

6z^{2} ni ikki tarafga qo’shing.

$$-5z^{2}-3z-11+6z^{2}=0$$

z^{2} ni olish uchun -5z^{2} va 6z^{2} ni birlashtirish.

$$z^{2}-3z-11=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 1 ni a, -3 ni b va -11 ni c bilan almashtiring.

$$z=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-11\right)}}{2}$$

-3 kvadratini chiqarish.

$$z=\frac{-\left(-3\right)±\sqrt{9-4\left(-11\right)}}{2}$$

-4 ni -11 marotabaga ko'paytirish.

$$z=\frac{-\left(-3\right)±\sqrt{9+44}}{2}$$

9 ni 44 ga qo'shish.

$$z=\frac{-\left(-3\right)±\sqrt{53}}{2}$$

-3 ning teskarisi 3 ga teng.

$$z=\frac{3±\sqrt{53}}{2}$$

z=\frac{3±\sqrt{53}}{2} tenglamasini yeching, bunda ± musbat. 3 ni \sqrt{53} ga qo'shish.

$$z=\frac{\sqrt{53}+3}{2}$$

z=\frac{3±\sqrt{53}}{2} tenglamasini yeching, bunda ± manfiy. 3 dan \sqrt{53} ni ayirish.

$$z=\frac{3-\sqrt{53}}{2}$$

Tenglama yechildi.

$$z=\frac{\sqrt{53}+3}{2} z=\frac{3-\sqrt{53}}{2}$$

Kvadratni yakunlash bo'yicha qadamlar

6z^{2} ni ikki tarafga qo’shing.

$$-5z^{2}-3z-11+6z^{2}=0$$

z^{2} ni olish uchun -5z^{2} va 6z^{2} ni birlashtirish.

$$z^{2}-3z-11=0$$

11 ni ikki tarafga qo’shing. Har qanday songa nolni qo‘shsangiz, o‘zi chiqadi.

$$z^{2}-3z=11$$

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

$$z^{2}-3z+\left(-\frac{3}{2}\right)^{2}=11+\left(-\frac{3}{2}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{3}{2} kvadratini chiqarish.

$$z^{2}-3z+\frac{9}{4}=11+\frac{9}{4}$$

11 ni \frac{9}{4} ga qo'shish.

$$z^{2}-3z+\frac{9}{4}=\frac{53}{4}$$

z^{2}-3z+\frac{9}{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.

$$\left(z-\frac{3}{2}\right)^{2}=\frac{53}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(z-\frac{3}{2}\right)^{2}}=\sqrt{\frac{53}{4}}$$

Qisqartirish.

$$z-\frac{3}{2}=\frac{\sqrt{53}}{2} z-\frac{3}{2}=-\frac{\sqrt{53}}{2}$$

\frac{3}{2} ni tenglamaning ikkala tarafiga qo'shish.

$$z=\frac{\sqrt{53}+3}{2} z=\frac{3-\sqrt{53}}{2}$$