3x+sqrt{6x+4}=38 eching

$$3x+ \sqrt{ 6x+4 } = 38$$

x uchun yechish

$$x=10$$

Yechim qadamlari

Tenglamaning ikkala tarafidan 3x ni ayirish.

$$\sqrt{6x+4}=38-3x$$

Tenglamaning ikkala taraf kvadratini chiqarish.

$$\left(\sqrt{6x+4}\right)^{2}=\left(38-3x\right)^{2}$$

2 daraja ko‘rsatkichini \sqrt{6x+4} ga hisoblang va 6x+4 ni qiymatni oling.

$$6x+4=\left(38-3x\right)^{2}$$

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(38-3x\right)^{2} kengaytirilishi uchun ishlating.

$$6x+4=1444-228x+9x^{2}$$

Ikkala tarafdan 1444 ni ayirish.

$$6x+4-1444=-228x+9x^{2}$$

-1440 olish uchun 4 dan 1444 ni ayirish.

$$6x-1440=-228x+9x^{2}$$

228x ni ikki tarafga qo’shing.

$$6x-1440+228x=9x^{2}$$

234x ni olish uchun 6x va 228x ni birlashtirish.

$$234x-1440=9x^{2}$$

Ikkala tarafdan 9x^{2} ni ayirish.

$$234x-1440-9x^{2}=0$$

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.

$$-9x^{2}+234x-1440=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} -9 ni a, 234 ni b va -1440 ni c bilan almashtiring.

$$x=\frac{-234±\sqrt{234^{2}-4\left(-9\right)\left(-1440\right)}}{2\left(-9\right)}$$

234 kvadratini chiqarish.

$$x=\frac{-234±\sqrt{54756-4\left(-9\right)\left(-1440\right)}}{2\left(-9\right)}$$

-4 ni -9 marotabaga ko'paytirish.

$$x=\frac{-234±\sqrt{54756+36\left(-1440\right)}}{2\left(-9\right)}$$

36 ni -1440 marotabaga ko'paytirish.

$$x=\frac{-234±\sqrt{54756-51840}}{2\left(-9\right)}$$

54756 ni -51840 ga qo'shish.

$$x=\frac{-234±\sqrt{2916}}{2\left(-9\right)}$$

2916 ning kvadrat ildizini chiqarish.

$$x=\frac{-234±54}{2\left(-9\right)}$$

2 ni -9 marotabaga ko'paytirish.

$$x=\frac{-234±54}{-18}$$

x=\frac{-234±54}{-18} tenglamasini yeching, bunda ± musbat. -234 ni 54 ga qo'shish.

$$x=-\frac{180}{-18}$$

-180 ni -18 ga bo'lish.

$$x=10$$

x=\frac{-234±54}{-18} tenglamasini yeching, bunda ± manfiy. -234 dan 54 ni ayirish.

$$x=-\frac{288}{-18}$$

-288 ni -18 ga bo'lish.

$$x=16$$

Tenglama yechildi.

$$x=10 x=16$$

3x+\sqrt{6x+4}=38 tenglamasida x uchun 10 ni almashtiring.

$$3\times 10+\sqrt{6\times 10+4}=38$$

Qisqartirish. x=10 tenglamani qoniqtiradi.

$$38=38$$

3x+\sqrt{6x+4}=38 tenglamasida x uchun 16 ni almashtiring.

$$3\times 16+\sqrt{6\times 16+4}=38$$

Qisqartirish. x=16 qiymati bu tenglamani qoniqtirmaydi.

$$58=38$$

\sqrt{6x+4}=38-3x tenglamasi noyob yechimga ega.

$$x=10$$