a^2-10a=4 eching

$$a ^ { 2 } - 10 a = 4$$

a uchun yechish

$$a=\sqrt{29}+5\approx 10,385164807$$

$$a=5-\sqrt{29}\approx -0,385164807$$

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.

$$a^{2}-10a=4$$

Tenglamaning ikkala tarafidan 4 ni ayirish.

$$a^{2}-10a-4=4-4$$

O‘zidan 4 ayirilsa 0 qoladi.

$$a^{2}-10a-4=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 1 ni a, -10 ni b va -4 ni c bilan almashtiring.

$$a=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\left(-4\right)}}{2}$$

-10 kvadratini chiqarish.

$$a=\frac{-\left(-10\right)±\sqrt{100-4\left(-4\right)}}{2}$$

-4 ni -4 marotabaga ko'paytirish.

$$a=\frac{-\left(-10\right)±\sqrt{100+16}}{2}$$

100 ni 16 ga qo'shish.

$$a=\frac{-\left(-10\right)±\sqrt{116}}{2}$$

116 ning kvadrat ildizini chiqarish.

$$a=\frac{-\left(-10\right)±2\sqrt{29}}{2}$$

-10 ning teskarisi 10 ga teng.

$$a=\frac{10±2\sqrt{29}}{2}$$

a=\frac{10±2\sqrt{29}}{2} tenglamasini yeching, bunda ± musbat. 10 ni 2\sqrt{29} ga qo'shish.

$$a=\frac{2\sqrt{29}+10}{2}$$

10+2\sqrt{29} ni 2 ga bo'lish.

$$a=\sqrt{29}+5$$

a=\frac{10±2\sqrt{29}}{2} tenglamasini yeching, bunda ± manfiy. 10 dan 2\sqrt{29} ni ayirish.

$$a=\frac{10-2\sqrt{29}}{2}$$

10-2\sqrt{29} ni 2 ga bo'lish.

$$a=5-\sqrt{29}$$

Tenglama yechildi.

$$a=\sqrt{29}+5 a=5-\sqrt{29}$$

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.

$$a^{2}-10a=4$$

-10 ni bo‘lish, x shartining koeffitsienti, 2 ga -5 olish uchun. Keyin, -5 ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$a^{2}-10a+\left(-5\right)^{2}=4+\left(-5\right)^{2}$$

-5 kvadratini chiqarish.

$$a^{2}-10a+25=4+25$$

4 ni 25 ga qo'shish.

$$a^{2}-10a+25=29$$

a^{2}-10a+25 omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(a-5\right)^{2}=29$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(a-5\right)^{2}}=\sqrt{29}$$

Qisqartirish.

$$a-5=\sqrt{29} a-5=-\sqrt{29}$$

5 ni tenglamaning ikkala tarafiga qo'shish.

$$a=\sqrt{29}+5 a=5-\sqrt{29}$$