5x^2-43x-125=7x eching

$$5 { x }^{ 2 } -43x-125=7x$$

x uchun yechish

$$x=5\sqrt{2}+5\approx 12,071067812$$

$$x=5-5\sqrt{2}\approx -2,071067812$$

Kvadrat tenglamalaridan foydalanish qoidalari

Ikkala tarafdan 7x ni ayirish.

$$5x^{2}-43x-125-7x=0$$

-50x ni olish uchun -43x va -7x ni birlashtirish.

$$5x^{2}-50x-125=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 5 ni a, -50 ni b va -125 ni c bilan almashtiring.

$$x=\frac{-\left(-50\right)±\sqrt{\left(-50\right)^{2}-4\times 5\left(-125\right)}}{2\times 5}$$

-50 kvadratini chiqarish.

$$x=\frac{-\left(-50\right)±\sqrt{2500-4\times 5\left(-125\right)}}{2\times 5}$$

-4 ni 5 marotabaga ko'paytirish.

$$x=\frac{-\left(-50\right)±\sqrt{2500-20\left(-125\right)}}{2\times 5}$$

-20 ni -125 marotabaga ko'paytirish.

$$x=\frac{-\left(-50\right)±\sqrt{2500+2500}}{2\times 5}$$

2500 ni 2500 ga qo'shish.

$$x=\frac{-\left(-50\right)±\sqrt{5000}}{2\times 5}$$

5000 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-50\right)±50\sqrt{2}}{2\times 5}$$

-50 ning teskarisi 50 ga teng.

$$x=\frac{50±50\sqrt{2}}{2\times 5}$$

2 ni 5 marotabaga ko'paytirish.

$$x=\frac{50±50\sqrt{2}}{10}$$

x=\frac{50±50\sqrt{2}}{10} tenglamasini yeching, bunda ± musbat. 50 ni 50\sqrt{2} ga qo'shish.

$$x=\frac{50\sqrt{2}+50}{10}$$

50+50\sqrt{2} ni 10 ga bo'lish.

$$x=5\sqrt{2}+5$$

x=\frac{50±50\sqrt{2}}{10} tenglamasini yeching, bunda ± manfiy. 50 dan 50\sqrt{2} ni ayirish.

$$x=\frac{50-50\sqrt{2}}{10}$$

50-50\sqrt{2} ni 10 ga bo'lish.

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

Tenglama yechildi.

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

Kvadratni yakunlash bo'yicha qadamlar

Ikkala tarafdan 7x ni ayirish.

$$5x^{2}-43x-125-7x=0$$

-50x ni olish uchun -43x va -7x ni birlashtirish.

$$5x^{2}-50x-125=0$$

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

$$5x^{2}-50x=125$$

Ikki tarafini 5 ga bo‘ling.

$$\frac{5x^{2}-50x}{5}=\frac{125}{5}$$

5 ga bo'lish 5 ga ko'paytirishni bekor qiladi.

$$x^{2}+\left(-\frac{50}{5}\right)x=\frac{125}{5}$$

-50 ni 5 ga bo'lish.

$$x^{2}-10x=\frac{125}{5}$$

125 ni 5 ga bo'lish.

$$x^{2}-10x=25$$

-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.

$$x^{2}-10x+\left(-5\right)^{2}=25+\left(-5\right)^{2}$$

-5 kvadratini chiqarish.

$$x^{2}-10x+25=25+25$$

25 ni 25 ga qo'shish.

$$x^{2}-10x+25=50$$

x^{2}-10x+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.

$$\left(x-5\right)^{2}=50$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-5\right)^{2}}=\sqrt{50}$$

Qisqartirish.

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

5 ni tenglamaning ikkala tarafiga qo'shish.

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