-x^2-5x+4=0 eching

$$- x ^ { 2 } - 5 x + 4 = 0$$

x uchun yechish

$$x=\frac{\sqrt{41}-5}{2}\approx 0,701562119$$

$$x=\frac{-\sqrt{41}-5}{2}\approx -5,701562119$$

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.

$$-x^{2}-5x+4=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} -1 ni a, -5 ni b va 4 ni c bilan almashtiring.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\left(-1\right)\times 4}}{2\left(-1\right)}$$

-5 kvadratini chiqarish.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\left(-1\right)\times 4}}{2\left(-1\right)}$$

-4 ni -1 marotabaga ko'paytirish.

$$x=\frac{-\left(-5\right)±\sqrt{25+4\times 4}}{2\left(-1\right)}$$

4 ni 4 marotabaga ko'paytirish.

$$x=\frac{-\left(-5\right)±\sqrt{25+16}}{2\left(-1\right)}$$

25 ni 16 ga qo'shish.

$$x=\frac{-\left(-5\right)±\sqrt{41}}{2\left(-1\right)}$$

-5 ning teskarisi 5 ga teng.

$$x=\frac{5±\sqrt{41}}{2\left(-1\right)}$$

2 ni -1 marotabaga ko'paytirish.

$$x=\frac{5±\sqrt{41}}{-2}$$

x=\frac{5±\sqrt{41}}{-2} tenglamasini yeching, bunda ± musbat. 5 ni \sqrt{41} ga qo'shish.

$$x=\frac{\sqrt{41}+5}{-2}$$

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

$$x=\frac{-\sqrt{41}-5}{2}$$

x=\frac{5±\sqrt{41}}{-2} tenglamasini yeching, bunda ± manfiy. 5 dan \sqrt{41} ni ayirish.

$$x=\frac{5-\sqrt{41}}{-2}$$

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

$$x=\frac{\sqrt{41}-5}{2}$$

Tenglama yechildi.

$$x=\frac{-\sqrt{41}-5}{2} x=\frac{\sqrt{41}-5}{2}$$

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.

$$-x^{2}-5x+4=0$$

Tenglamaning ikkala tarafidan 4 ni ayirish.

$$-x^{2}-5x+4-4=-4$$

O‘zidan 4 ayirilsa 0 qoladi.

$$-x^{2}-5x=-4$$

Ikki tarafini -1 ga bo‘ling.

$$\frac{-x^{2}-5x}{-1}=-\frac{4}{-1}$$

-1 ga bo'lish -1 ga ko'paytirishni bekor qiladi.

$$x^{2}+\left(-\frac{5}{-1}\right)x=-\frac{4}{-1}$$

-5 ni -1 ga bo'lish.

$$x^{2}+5x=-\frac{4}{-1}$$

-4 ni -1 ga bo'lish.

$$x^{2}+5x=4$$

5 ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{5}{2} olish uchun. Keyin, \frac{5}{2} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+5x+\left(\frac{5}{2}\right)^{2}=4+\left(\frac{5}{2}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{5}{2} kvadratini chiqarish.

$$x^{2}+5x+\frac{25}{4}=4+\frac{25}{4}$$

4 ni \frac{25}{4} ga qo'shish.

$$x^{2}+5x+\frac{25}{4}=\frac{41}{4}$$

x^{2}+5x+\frac{25}{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.

$$\left(x+\frac{5}{2}\right)^{2}=\frac{41}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{5}{2}\right)^{2}}=\sqrt{\frac{41}{4}}$$

Qisqartirish.

$$x+\frac{5}{2}=\frac{\sqrt{41}}{2} x+\frac{5}{2}=-\frac{\sqrt{41}}{2}$$

Tenglamaning ikkala tarafidan \frac{5}{2} ni ayirish.

$$x=\frac{\sqrt{41}-5}{2} x=\frac{-\sqrt{41}-5}{2}$$