1=x^2-8x+15 eching

$$1 = x ^ { 2 } - 8 x + 15$$

x uchun yechish

$$x=\sqrt{2}+4\approx 5,414213562$$

$$x=4-\sqrt{2}\approx 2,585786438$$

Kvadrat tenglamalaridan foydalanish qoidalari

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$x^{2}-8x+15=1$$

Ikkala tarafdan 1 ni ayirish.

$$x^{2}-8x+15-1=0$$

14 olish uchun 15 dan 1 ni ayirish.

$$x^{2}-8x+14=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 1 ni a, -8 ni b va 14 ni c bilan almashtiring.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 14}}{2}$$

-8 kvadratini chiqarish.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 14}}{2}$$

-4 ni 14 marotabaga ko'paytirish.

$$x=\frac{-\left(-8\right)±\sqrt{64-56}}{2}$$

64 ni -56 ga qo'shish.

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

8 ning kvadrat ildizini chiqarish.

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

-8 ning teskarisi 8 ga teng.

$$x=\frac{8±2\sqrt{2}}{2}$$

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

$$x=\frac{2\sqrt{2}+8}{2}$$

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

$$x=\sqrt{2}+4$$

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

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

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

$$x=4-\sqrt{2}$$

Tenglama yechildi.

$$x=\sqrt{2}+4 x=4-\sqrt{2}$$

Kvadratni yakunlash bo'yicha qadamlar

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$x^{2}-8x+15=1$$

Ikkala tarafdan 15 ni ayirish.

$$x^{2}-8x=1-15$$

-14 olish uchun 1 dan 15 ni ayirish.

$$x^{2}-8x=-14$$

-8 ni bo‘lish, x shartining koeffitsienti, 2 ga -4 olish uchun. Keyin, -4 ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}-8x+\left(-4\right)^{2}=-14+\left(-4\right)^{2}$$

-4 kvadratini chiqarish.

$$x^{2}-8x+16=-14+16$$

-14 ni 16 ga qo'shish.

$$x^{2}-8x+16=2$$

x^{2}-8x+16 omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x-4\right)^{2}=2$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-4\right)^{2}}=\sqrt{2}$$

Qisqartirish.

$$x-4=\sqrt{2} x-4=-\sqrt{2}$$

4 ni tenglamaning ikkala tarafiga qo'shish.

$$x=\sqrt{2}+4 x=4-\sqrt{2}$$