a^2+a-5=0 eching

$$a ^ { 2 } + a - 5 = 0$$

a uchun yechish

$$a = \frac{\sqrt{21} - 1}{2} \approx 1,791287847$$

$$a=\frac{-\sqrt{21}-1}{2}\approx -2,791287847$$

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.

$$a^{2}+a-5=0$$

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

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

1 kvadratini chiqarish.

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

-4 ni -5 marotabaga ko'paytirish.

$$a=\frac{-1±\sqrt{1+20}}{2}$$

1 ni 20 ga qo'shish.

$$a=\frac{-1±\sqrt{21}}{2}$$

a=\frac{-1±\sqrt{21}}{2} tenglamasini yeching, bunda ± musbat. -1 ni \sqrt{21} ga qo'shish.

$$a=\frac{\sqrt{21}-1}{2}$$

a=\frac{-1±\sqrt{21}}{2} tenglamasini yeching, bunda ± manfiy. -1 dan \sqrt{21} ni ayirish.

$$a=\frac{-\sqrt{21}-1}{2}$$

Tenglama yechildi.

$$a=\frac{\sqrt{21}-1}{2} a=\frac{-\sqrt{21}-1}{2}$$

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.

$$a^{2}+a-5=0$$

5 ni tenglamaning ikkala tarafiga qo'shish.

$$a^{2}+a-5-\left(-5\right)=-\left(-5\right)$$

O‘zidan -5 ayirilsa 0 qoladi.

$$a^{2}+a=-\left(-5\right)$$

0 dan -5 ni ayirish.

$$a^{2}+a=5$$

1 ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{1}{2} olish uchun. Keyin, \frac{1}{2} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$a^{2}+a+\left(\frac{1}{2}\right)^{2}=5+\left(\frac{1}{2}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{1}{2} kvadratini chiqarish.

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

5 ni \frac{1}{4} ga qo'shish.

$$a^{2}+a+\frac{1}{4}=\frac{21}{4}$$

a^{2}+a+\frac{1}{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.

$$\left(a+\frac{1}{2}\right)^{2}=\frac{21}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(a+\frac{1}{2}\right)^{2}}=\sqrt{\frac{21}{4}}$$

Qisqartirish.

$$a+\frac{1}{2}=\frac{\sqrt{21}}{2} a+\frac{1}{2}=-\frac{\sqrt{21}}{2}$$

Tenglamaning ikkala tarafidan \frac{1}{2} ni ayirish.

$$a=\frac{\sqrt{21}-1}{2} a=\frac{-\sqrt{21}-1}{2}$$