3:x+4=x+6:x+14 eching

$$3 : x + 4 = x + 6 : x + 14$$

x uchun yechish (complex solution)

$$x=\sqrt{22}-5\approx -0,30958424$$

$$x=-\left(\sqrt{22}+5\right)\approx -9,69041576$$

Kvadrat tenglamalaridan foydalanish qoidalari

x qiymati 0 teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini x ga ko'paytirish.

$$3+x\times 4=xx+6+x\times 14$$

x^{2} hosil qilish uchun x va x ni ko'paytirish.

$$3+x\times 4=x^{2}+6+x\times 14$$

Ikkala tarafdan x^{2} ni ayirish.

$$3+x\times 4-x^{2}=6+x\times 14$$

Ikkala tarafdan 6 ni ayirish.

$$3+x\times 4-x^{2}-6=x\times 14$$

-3 olish uchun 3 dan 6 ni ayirish.

$$-3+x\times 4-x^{2}=x\times 14$$

Ikkala tarafdan x\times 14 ni ayirish.

$$-3+x\times 4-x^{2}-x\times 14=0$$

-10x ni olish uchun x\times 4 va -x\times 14 ni birlashtirish.

$$-3-10x-x^{2}=0$$

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.

$$-x^{2}-10x-3=0$$

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 -3 ni c bilan almashtiring.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\left(-1\right)\left(-3\right)}}{2\left(-1\right)}$$

-10 kvadratini chiqarish.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\left(-1\right)\left(-3\right)}}{2\left(-1\right)}$$

-4 ni -1 marotabaga ko'paytirish.

$$x=\frac{-\left(-10\right)±\sqrt{100+4\left(-3\right)}}{2\left(-1\right)}$$

4 ni -3 marotabaga ko'paytirish.

$$x=\frac{-\left(-10\right)±\sqrt{100-12}}{2\left(-1\right)}$$

100 ni -12 ga qo'shish.

$$x=\frac{-\left(-10\right)±\sqrt{88}}{2\left(-1\right)}$$

88 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-10\right)±2\sqrt{22}}{2\left(-1\right)}$$

-10 ning teskarisi 10 ga teng.

$$x=\frac{10±2\sqrt{22}}{2\left(-1\right)}$$

2 ni -1 marotabaga ko'paytirish.

$$x=\frac{10±2\sqrt{22}}{-2}$$

x=\frac{10±2\sqrt{22}}{-2} tenglamasini yeching, bunda ± musbat. 10 ni 2\sqrt{22} ga qo'shish.

$$x=\frac{2\sqrt{22}+10}{-2}$$

10+2\sqrt{22} ni -2 ga bo'lish.

$$x=-\left(\sqrt{22}+5\right)$$

x=\frac{10±2\sqrt{22}}{-2} tenglamasini yeching, bunda ± manfiy. 10 dan 2\sqrt{22} ni ayirish.

$$x=\frac{10-2\sqrt{22}}{-2}$$

10-2\sqrt{22} ni -2 ga bo'lish.

$$x=\sqrt{22}-5$$

Tenglama yechildi.

$$x=-\left(\sqrt{22}+5\right) x=\sqrt{22}-5$$

Kvadratni yakunlash bo'yicha qadamlar

x qiymati 0 teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini x ga ko'paytirish.

$$3+x\times 4=xx+6+x\times 14$$

x^{2} hosil qilish uchun x va x ni ko'paytirish.

$$3+x\times 4=x^{2}+6+x\times 14$$

Ikkala tarafdan x^{2} ni ayirish.

$$3+x\times 4-x^{2}=6+x\times 14$$

Ikkala tarafdan x\times 14 ni ayirish.

$$3+x\times 4-x^{2}-x\times 14=6$$

-10x ni olish uchun x\times 4 va -x\times 14 ni birlashtirish.

$$3-10x-x^{2}=6$$

Ikkala tarafdan 3 ni ayirish.

$$-10x-x^{2}=6-3$$

3 olish uchun 6 dan 3 ni ayirish.

$$-10x-x^{2}=3$$

Bu kabi kvadrat tenglamalarni kvadratni yakunlab yechish mumkin. Kvadratni yechish uchun tenglama avval ushbu shaklda bo'lishi shart: x^{2}+bx=c.

$$-x^{2}-10x=3$$

Ikki tarafini -1 ga bo‘ling.

$$\frac{-x^{2}-10x}{-1}=\frac{3}{-1}$$

-1 ga bo'lish -1 ga ko'paytirishni bekor qiladi.

$$x^{2}+\left(-\frac{10}{-1}\right)x=\frac{3}{-1}$$

-10 ni -1 ga bo'lish.

$$x^{2}+10x=\frac{3}{-1}$$

3 ni -1 ga bo'lish.

$$x^{2}+10x=-3$$

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.

$$x^{2}+10x+5^{2}=-3+5^{2}$$

5 kvadratini chiqarish.

$$x^{2}+10x+25=-3+25$$

-3 ni 25 ga qo'shish.

$$x^{2}+10x+25=22$$

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.

$$\left(x+5\right)^{2}=22$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+5\right)^{2}}=\sqrt{22}$$

Qisqartirish.

$$x+5=\sqrt{22} x+5=-\sqrt{22}$$

Tenglamaning ikkala tarafidan 5 ni ayirish.

$$x=\sqrt{22}-5 x=-\sqrt{22}-5$$

x uchun yechish

$$x=\sqrt{22}-5\approx -0,30958424$$

$$x=-\sqrt{22}-5\approx -9,69041576$$

Kvadrat tenglamalaridan foydalanish qoidalari

x qiymati 0 teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini x ga ko'paytirish.

$$3+x\times 4=xx+6+x\times 14$$

x^{2} hosil qilish uchun x va x ni ko'paytirish.

$$3+x\times 4=x^{2}+6+x\times 14$$

Ikkala tarafdan x^{2} ni ayirish.

$$3+x\times 4-x^{2}=6+x\times 14$$

Ikkala tarafdan 6 ni ayirish.

$$3+x\times 4-x^{2}-6=x\times 14$$

-3 olish uchun 3 dan 6 ni ayirish.

$$-3+x\times 4-x^{2}=x\times 14$$

Ikkala tarafdan x\times 14 ni ayirish.

$$-3+x\times 4-x^{2}-x\times 14=0$$

-10x ni olish uchun x\times 4 va -x\times 14 ni birlashtirish.

$$-3-10x-x^{2}=0$$

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.

$$-x^{2}-10x-3=0$$

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 -3 ni c bilan almashtiring.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\left(-1\right)\left(-3\right)}}{2\left(-1\right)}$$

-10 kvadratini chiqarish.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\left(-1\right)\left(-3\right)}}{2\left(-1\right)}$$

-4 ni -1 marotabaga ko'paytirish.

$$x=\frac{-\left(-10\right)±\sqrt{100+4\left(-3\right)}}{2\left(-1\right)}$$

4 ni -3 marotabaga ko'paytirish.

$$x=\frac{-\left(-10\right)±\sqrt{100-12}}{2\left(-1\right)}$$

100 ni -12 ga qo'shish.

$$x=\frac{-\left(-10\right)±\sqrt{88}}{2\left(-1\right)}$$

88 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-10\right)±2\sqrt{22}}{2\left(-1\right)}$$

-10 ning teskarisi 10 ga teng.

$$x=\frac{10±2\sqrt{22}}{2\left(-1\right)}$$

2 ni -1 marotabaga ko'paytirish.

$$x=\frac{10±2\sqrt{22}}{-2}$$

x=\frac{10±2\sqrt{22}}{-2} tenglamasini yeching, bunda ± musbat. 10 ni 2\sqrt{22} ga qo'shish.

$$x=\frac{2\sqrt{22}+10}{-2}$$

10+2\sqrt{22} ni -2 ga bo'lish.

$$x=-\left(\sqrt{22}+5\right)$$

x=\frac{10±2\sqrt{22}}{-2} tenglamasini yeching, bunda ± manfiy. 10 dan 2\sqrt{22} ni ayirish.

$$x=\frac{10-2\sqrt{22}}{-2}$$

10-2\sqrt{22} ni -2 ga bo'lish.

$$x=\sqrt{22}-5$$

Tenglama yechildi.

$$x=-\left(\sqrt{22}+5\right) x=\sqrt{22}-5$$

Kvadratni yakunlash bo'yicha qadamlar

x qiymati 0 teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini x ga ko'paytirish.

$$3+x\times 4=xx+6+x\times 14$$

x^{2} hosil qilish uchun x va x ni ko'paytirish.

$$3+x\times 4=x^{2}+6+x\times 14$$

Ikkala tarafdan x^{2} ni ayirish.

$$3+x\times 4-x^{2}=6+x\times 14$$

Ikkala tarafdan x\times 14 ni ayirish.

$$3+x\times 4-x^{2}-x\times 14=6$$

-10x ni olish uchun x\times 4 va -x\times 14 ni birlashtirish.

$$3-10x-x^{2}=6$$

Ikkala tarafdan 3 ni ayirish.

$$-10x-x^{2}=6-3$$

3 olish uchun 6 dan 3 ni ayirish.

$$-10x-x^{2}=3$$

Bu kabi kvadrat tenglamalarni kvadratni yakunlab yechish mumkin. Kvadratni yechish uchun tenglama avval ushbu shaklda bo'lishi shart: x^{2}+bx=c.

$$-x^{2}-10x=3$$

Ikki tarafini -1 ga bo‘ling.

$$\frac{-x^{2}-10x}{-1}=\frac{3}{-1}$$

-1 ga bo'lish -1 ga ko'paytirishni bekor qiladi.

$$x^{2}+\left(-\frac{10}{-1}\right)x=\frac{3}{-1}$$

-10 ni -1 ga bo'lish.

$$x^{2}+10x=\frac{3}{-1}$$

3 ni -1 ga bo'lish.

$$x^{2}+10x=-3$$

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.

$$x^{2}+10x+5^{2}=-3+5^{2}$$

5 kvadratini chiqarish.

$$x^{2}+10x+25=-3+25$$

-3 ni 25 ga qo'shish.

$$x^{2}+10x+25=22$$

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.

$$\left(x+5\right)^{2}=22$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+5\right)^{2}}=\sqrt{22}$$

Qisqartirish.

$$x+5=\sqrt{22} x+5=-\sqrt{22}$$

Tenglamaning ikkala tarafidan 5 ni ayirish.

$$x=\sqrt{22}-5 x=-\sqrt{22}-5$$