0=x^2+11x-8 eching

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

x uchun yechish

$$x=\frac{3\sqrt{17}-11}{2}\approx 0,684658438$$

$$x=\frac{-3\sqrt{17}-11}{2}\approx -11,684658438$$

Kvadrat tenglamalaridan foydalanish qoidalari

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

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

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 -8 ni c bilan almashtiring.

$$x=\frac{-11±\sqrt{11^{2}-4\left(-8\right)}}{2}$$

11 kvadratini chiqarish.

$$x=\frac{-11±\sqrt{121-4\left(-8\right)}}{2}$$

-4 ni -8 marotabaga ko'paytirish.

$$x=\frac{-11±\sqrt{121+32}}{2}$$

121 ni 32 ga qo'shish.

$$x=\frac{-11±\sqrt{153}}{2}$$

153 ning kvadrat ildizini chiqarish.

$$x=\frac{-11±3\sqrt{17}}{2}$$

x=\frac{-11±3\sqrt{17}}{2} tenglamasini yeching, bunda ± musbat. -11 ni 3\sqrt{17} ga qo'shish.

$$x=\frac{3\sqrt{17}-11}{2}$$

x=\frac{-11±3\sqrt{17}}{2} tenglamasini yeching, bunda ± manfiy. -11 dan 3\sqrt{17} ni ayirish.

$$x=\frac{-3\sqrt{17}-11}{2}$$

Tenglama yechildi.

$$x=\frac{3\sqrt{17}-11}{2} x=\frac{-3\sqrt{17}-11}{2}$$

Kvadratni yakunlash bo'yicha qadamlar

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

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

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

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

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.

$$x^{2}+11x+\left(\frac{11}{2}\right)^{2}=8+\left(\frac{11}{2}\right)^{2}$$

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

$$x^{2}+11x+\frac{121}{4}=8+\frac{121}{4}$$

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

$$x^{2}+11x+\frac{121}{4}=\frac{153}{4}$$

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.

$$\left(x+\frac{11}{2}\right)^{2}=\frac{153}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{11}{2}\right)^{2}}=\sqrt{\frac{153}{4}}$$

Qisqartirish.

$$x+\frac{11}{2}=\frac{3\sqrt{17}}{2} x+\frac{11}{2}=-\frac{3\sqrt{17}}{2}$$

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

$$x=\frac{3\sqrt{17}-11}{2} x=\frac{-3\sqrt{17}-11}{2}$$