(x)=-x^2+4x+6 eching

$$( x ) = - x ^ { 2 } + 4 x + 6$$

x uchun yechish

$$x = \frac{\sqrt{33} + 3}{2} \approx 4,372281323$$

$$x=\frac{3-\sqrt{33}}{2}\approx -1,372281323$$

Kvadrat tenglamalaridan foydalanish qoidalari

x^{2} ni ikki tarafga qo’shing.

$$x+x^{2}=4x+6$$

Ikkala tarafdan 4x ni ayirish.

$$x+x^{2}-4x=6$$

-3x ni olish uchun x va -4x ni birlashtirish.

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

Ikkala tarafdan 6 ni ayirish.

$$-3x+x^{2}-6=0$$

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.

$$x^{2}-3x-6=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 1 ni a, -3 ni b va -6 ni c bilan almashtiring.

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

-3 kvadratini chiqarish.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-6\right)}}{2}$$

-4 ni -6 marotabaga ko'paytirish.

$$x=\frac{-\left(-3\right)±\sqrt{9+24}}{2}$$

9 ni 24 ga qo'shish.

$$x=\frac{-\left(-3\right)±\sqrt{33}}{2}$$

-3 ning teskarisi 3 ga teng.

$$x=\frac{3±\sqrt{33}}{2}$$

x=\frac{3±\sqrt{33}}{2} tenglamasini yeching, bunda ± musbat. 3 ni \sqrt{33} ga qo'shish.

$$x=\frac{\sqrt{33}+3}{2}$$

x=\frac{3±\sqrt{33}}{2} tenglamasini yeching, bunda ± manfiy. 3 dan \sqrt{33} ni ayirish.

$$x=\frac{3-\sqrt{33}}{2}$$

Tenglama yechildi.

$$x=\frac{\sqrt{33}+3}{2} x=\frac{3-\sqrt{33}}{2}$$

Kvadratni yakunlash bo'yicha qadamlar

x^{2} ni ikki tarafga qo’shing.

$$x+x^{2}=4x+6$$

Ikkala tarafdan 4x ni ayirish.

$$x+x^{2}-4x=6$$

-3x ni olish uchun x va -4x ni birlashtirish.

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

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

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

-3 ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{3}{2} olish uchun. Keyin, -\frac{3}{2} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

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

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{3}{2} kvadratini chiqarish.

$$x^{2}-3x+\frac{9}{4}=6+\frac{9}{4}$$

6 ni \frac{9}{4} ga qo'shish.

$$x^{2}-3x+\frac{9}{4}=\frac{33}{4}$$

x^{2}-3x+\frac{9}{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.

$$\left(x-\frac{3}{2}\right)^{2}=\frac{33}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{33}{4}}$$

Qisqartirish.

$$x-\frac{3}{2}=\frac{\sqrt{33}}{2} x-\frac{3}{2}=-\frac{\sqrt{33}}{2}$$

\frac{3}{2} ni tenglamaning ikkala tarafiga qo'shish.

$$x=\frac{\sqrt{33}+3}{2} x=\frac{3-\sqrt{33}}{2}$$