4x^2+7x-8=x eching

$$4 x ^ { 2 } + 7 x - 8 = x$$

x uchun yechish

$$x=\frac{\sqrt{41}-3}{4}\approx 0,850781059$$

$$x=\frac{-\sqrt{41}-3}{4}\approx -2,350781059$$

Kvadrat tenglamalaridan foydalanish qoidalari

Ikkala tarafdan x ni ayirish.

$$4x^{2}+7x-8-x=0$$

6x ni olish uchun 7x va -x ni birlashtirish.

$$4x^{2}+6x-8=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 4 ni a, 6 ni b va -8 ni c bilan almashtiring.

$$x=\frac{-6±\sqrt{6^{2}-4\times 4\left(-8\right)}}{2\times 4}$$

6 kvadratini chiqarish.

$$x=\frac{-6±\sqrt{36-4\times 4\left(-8\right)}}{2\times 4}$$

-4 ni 4 marotabaga ko'paytirish.

$$x=\frac{-6±\sqrt{36-16\left(-8\right)}}{2\times 4}$$

-16 ni -8 marotabaga ko'paytirish.

$$x=\frac{-6±\sqrt{36+128}}{2\times 4}$$

36 ni 128 ga qo'shish.

$$x=\frac{-6±\sqrt{164}}{2\times 4}$$

164 ning kvadrat ildizini chiqarish.

$$x=\frac{-6±2\sqrt{41}}{2\times 4}$$

2 ni 4 marotabaga ko'paytirish.

$$x=\frac{-6±2\sqrt{41}}{8}$$

x=\frac{-6±2\sqrt{41}}{8} tenglamasini yeching, bunda ± musbat. -6 ni 2\sqrt{41} ga qo'shish.

$$x=\frac{2\sqrt{41}-6}{8}$$

-6+2\sqrt{41} ni 8 ga bo'lish.

$$x=\frac{\sqrt{41}-3}{4}$$

x=\frac{-6±2\sqrt{41}}{8} tenglamasini yeching, bunda ± manfiy. -6 dan 2\sqrt{41} ni ayirish.

$$x=\frac{-2\sqrt{41}-6}{8}$$

-6-2\sqrt{41} ni 8 ga bo'lish.

$$x=\frac{-\sqrt{41}-3}{4}$$

Tenglama yechildi.

$$x=\frac{\sqrt{41}-3}{4} x=\frac{-\sqrt{41}-3}{4}$$

Kvadratni yakunlash bo'yicha qadamlar

Ikkala tarafdan x ni ayirish.

$$4x^{2}+7x-8-x=0$$

6x ni olish uchun 7x va -x ni birlashtirish.

$$4x^{2}+6x-8=0$$

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

$$4x^{2}+6x=8$$

Ikki tarafini 4 ga bo‘ling.

$$\frac{4x^{2}+6x}{4}=\frac{8}{4}$$

4 ga bo'lish 4 ga ko'paytirishni bekor qiladi.

$$x^{2}+\frac{6}{4}x=\frac{8}{4}$$

\frac{6}{4} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x^{2}+\frac{3}{2}x=\frac{8}{4}$$

8 ni 4 ga bo'lish.

$$x^{2}+\frac{3}{2}x=2$$

\frac{3}{2} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{3}{4} olish uchun. Keyin, \frac{3}{4} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+\frac{3}{2}x+\left(\frac{3}{4}\right)^{2}=2+\left(\frac{3}{4}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{3}{4} kvadratini chiqarish.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=2+\frac{9}{16}$$

2 ni \frac{9}{16} ga qo'shish.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=\frac{41}{16}$$

x^{2}+\frac{3}{2}x+\frac{9}{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.

$$\left(x+\frac{3}{4}\right)^{2}=\frac{41}{16}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{3}{4}\right)^{2}}=\sqrt{\frac{41}{16}}$$

Qisqartirish.

$$x+\frac{3}{4}=\frac{\sqrt{41}}{4} x+\frac{3}{4}=-\frac{\sqrt{41}}{4}$$

Tenglamaning ikkala tarafidan \frac{3}{4} ni ayirish.

$$x=\frac{\sqrt{41}-3}{4} x=\frac{-\sqrt{41}-3}{4}$$