x+3/2=5/x+4 eching

$$\frac { x + 3 } { 2 } = \frac { 5 } { x + 4 }$$

x uchun yechish

$$x=\frac{\sqrt{41}-7}{2}\approx -0,298437881$$

$$x=\frac{-\sqrt{41}-7}{2}\approx -6,701562119$$

Kvadrat tenglamalaridan foydalanish qoidalari

x qiymati -4 teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini 2\left(x+4\right) ga, 2,x+4 ning eng kichik karralisiga ko‘paytiring.

$$\left(x+4\right)\left(x+3\right)=2\times 5$$

x+4 ga x+3 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

$$x^{2}+7x+12=2\times 5$$

10 hosil qilish uchun 2 va 5 ni ko'paytirish.

$$x^{2}+7x+12=10$$

Ikkala tarafdan 10 ni ayirish.

$$x^{2}+7x+12-10=0$$

2 olish uchun 12 dan 10 ni ayirish.

$$x^{2}+7x+2=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 1 ni a, 7 ni b va 2 ni c bilan almashtiring.

$$x=\frac{-7±\sqrt{7^{2}-4\times 2}}{2}$$

7 kvadratini chiqarish.

$$x=\frac{-7±\sqrt{49-4\times 2}}{2}$$

-4 ni 2 marotabaga ko'paytirish.

$$x=\frac{-7±\sqrt{49-8}}{2}$$

49 ni -8 ga qo'shish.

$$x=\frac{-7±\sqrt{41}}{2}$$

x=\frac{-7±\sqrt{41}}{2} tenglamasini yeching, bunda ± musbat. -7 ni \sqrt{41} ga qo'shish.

$$x=\frac{\sqrt{41}-7}{2}$$

x=\frac{-7±\sqrt{41}}{2} tenglamasini yeching, bunda ± manfiy. -7 dan \sqrt{41} ni ayirish.

$$x=\frac{-\sqrt{41}-7}{2}$$

Tenglama yechildi.

$$x=\frac{\sqrt{41}-7}{2} x=\frac{-\sqrt{41}-7}{2}$$

Kvadratni yakunlash bo'yicha qadamlar

x qiymati -4 teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini 2\left(x+4\right) ga, 2,x+4 ning eng kichik karralisiga ko‘paytiring.

$$\left(x+4\right)\left(x+3\right)=2\times 5$$

x+4 ga x+3 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

$$x^{2}+7x+12=2\times 5$$

10 hosil qilish uchun 2 va 5 ni ko'paytirish.

$$x^{2}+7x+12=10$$

Ikkala tarafdan 12 ni ayirish.

$$x^{2}+7x=10-12$$

-2 olish uchun 10 dan 12 ni ayirish.

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

7 ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{7}{2} olish uchun. Keyin, \frac{7}{2} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+7x+\left(\frac{7}{2}\right)^{2}=-2+\left(\frac{7}{2}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{7}{2} kvadratini chiqarish.

$$x^{2}+7x+\frac{49}{4}=-2+\frac{49}{4}$$

-2 ni \frac{49}{4} ga qo'shish.

$$x^{2}+7x+\frac{49}{4}=\frac{41}{4}$$

x^{2}+7x+\frac{49}{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.

$$\left(x+\frac{7}{2}\right)^{2}=\frac{41}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{7}{2}\right)^{2}}=\sqrt{\frac{41}{4}}$$

Qisqartirish.

$$x+\frac{7}{2}=\frac{\sqrt{41}}{2} x+\frac{7}{2}=-\frac{\sqrt{41}}{2}$$

Tenglamaning ikkala tarafidan \frac{7}{2} ni ayirish.

$$x=\frac{\sqrt{41}-7}{2} x=\frac{-\sqrt{41}-7}{2}$$