x^2+18x-95=0 eching

$${ x }^{ 2 } +18x-95=0$$

x uchun yechish

$$x=4\sqrt{11}-9\approx 4,266499161$$

$$x=-4\sqrt{11}-9\approx -22,266499161$$

Kvadrat tenglamalaridan foydalanish qoidalari

ax^{2}+bx+c=0 shaklidagi barcha tenglamalarni kvadrat formulasi bilan yechish mumkin: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Kvadrat formula ikki yechmni taqdim qiladi, biri ± qo'shish bo'lganda, va ikkinchisi ayiruv bo'lganda.

$$x^{2}+18x-95=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 1 ni a, 18 ni b va -95 ni c bilan almashtiring.

$$x=\frac{-18±\sqrt{18^{2}-4\left(-95\right)}}{2}$$

18 kvadratini chiqarish.

$$x=\frac{-18±\sqrt{324-4\left(-95\right)}}{2}$$

-4 ni -95 marotabaga ko'paytirish.

$$x=\frac{-18±\sqrt{324+380}}{2}$$

324 ni 380 ga qo'shish.

$$x=\frac{-18±\sqrt{704}}{2}$$

704 ning kvadrat ildizini chiqarish.

$$x=\frac{-18±8\sqrt{11}}{2}$$

x=\frac{-18±8\sqrt{11}}{2} tenglamasini yeching, bunda ± musbat. -18 ni 8\sqrt{11} ga qo'shish.

$$x=\frac{8\sqrt{11}-18}{2}$$

-18+8\sqrt{11} ni 2 ga bo'lish.

$$x=4\sqrt{11}-9$$

x=\frac{-18±8\sqrt{11}}{2} tenglamasini yeching, bunda ± manfiy. -18 dan 8\sqrt{11} ni ayirish.

$$x=\frac{-8\sqrt{11}-18}{2}$$

-18-8\sqrt{11} ni 2 ga bo'lish.

$$x=-4\sqrt{11}-9$$

Tenglama yechildi.

$$x=4\sqrt{11}-9 x=-4\sqrt{11}-9$$

Kvadratni yakunlash bo'yicha qadamlar

Bu kabi kvadrat tenglamalarni kvadratni yakunlab yechish mumkin. Kvadratni yechish uchun tenglama avval ushbu shaklda bo'lishi shart: x^{2}+bx=c.

$$x^{2}+18x-95=0$$

95 ni tenglamaning ikkala tarafiga qo'shish.

$$x^{2}+18x-95-\left(-95\right)=-\left(-95\right)$$

O‘zidan -95 ayirilsa 0 qoladi.

$$x^{2}+18x=-\left(-95\right)$$

0 dan -95 ni ayirish.

$$x^{2}+18x=95$$

18 ni bo‘lish, x shartining koeffitsienti, 2 ga 9 olish uchun. Keyin, 9 ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+18x+9^{2}=95+9^{2}$$

9 kvadratini chiqarish.

$$x^{2}+18x+81=95+81$$

95 ni 81 ga qo'shish.

$$x^{2}+18x+81=176$$

x^{2}+18x+81 omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+9\right)^{2}=176$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+9\right)^{2}}=\sqrt{176}$$

Qisqartirish.

$$x+9=4\sqrt{11} x+9=-4\sqrt{11}$$

Tenglamaning ikkala tarafidan 9 ni ayirish.

$$x=4\sqrt{11}-9 x=-4\sqrt{11}-9$$