0=-1/4x^2+3/2x+4 eching

$$0 = - \frac { 1 } { 4 } x ^ { 2 } + \frac { 3 } { 2 } x + 4$$

x uchun yechish

$$x=-2$$

$$x=8$$

Kvadrat tenglamalaridan foydalanish qoidalari

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$-\frac{1}{4}x^{2}+\frac{3}{2}x+4=0$$

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

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

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

$$x=\frac{-\frac{3}{2}±\sqrt{\frac{9}{4}-4\left(-\frac{1}{4}\right)\times 4}}{2\left(-\frac{1}{4}\right)}$$

-4 ni -\frac{1}{4} marotabaga ko'paytirish.

$$x=\frac{-\frac{3}{2}±\sqrt{\frac{9}{4}+4}}{2\left(-\frac{1}{4}\right)}$$

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

$$x=\frac{-\frac{3}{2}±\sqrt{\frac{25}{4}}}{2\left(-\frac{1}{4}\right)}$$

\frac{25}{4} ning kvadrat ildizini chiqarish.

$$x=\frac{-\frac{3}{2}±\frac{5}{2}}{2\left(-\frac{1}{4}\right)}$$

2 ni -\frac{1}{4} marotabaga ko'paytirish.

$$x=\frac{-\frac{3}{2}±\frac{5}{2}}{-\frac{1}{2}}$$

x=\frac{-\frac{3}{2}±\frac{5}{2}}{-\frac{1}{2}} tenglamasini yeching, bunda ± musbat. Umumiy maxrajni topib va hisoblovchini qo'shish orqali -\frac{3}{2} ni \frac{5}{2} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x=\frac{1}{-\frac{1}{2}}$$

1 ni -\frac{1}{2} ga bo'lish 1 ga k'paytirish -\frac{1}{2} ga qaytarish.

$$x=-2$$

x=\frac{-\frac{3}{2}±\frac{5}{2}}{-\frac{1}{2}} tenglamasini yeching, bunda ± manfiy. Umumiy maxrajni topib va suratlarni ayirib \frac{5}{2} ni -\frac{3}{2} dan ayirish. So'ngra imkoni boricha kasrni eng kichik shartga qisqartirish.

$$x=-\frac{4}{-\frac{1}{2}}$$

-4 ni -\frac{1}{2} ga bo'lish -4 ga k'paytirish -\frac{1}{2} ga qaytarish.

$$x=8$$

Tenglama yechildi.

$$x=-2 x=8$$

Kvadratni yakunlash bo'yicha qadamlar

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$-\frac{1}{4}x^{2}+\frac{3}{2}x+4=0$$

Ikkala tarafdan 4 ni ayirish. Har qanday sonni noldan ayirsangiz, o‘zining manfiyi chiqadi.

$$-\frac{1}{4}x^{2}+\frac{3}{2}x=-4$$

Ikkala tarafini -4 ga ko‘paytiring.

$$\frac{-\frac{1}{4}x^{2}+\frac{3}{2}x}{-\frac{1}{4}}=-\frac{4}{-\frac{1}{4}}$$

-\frac{1}{4} ga bo'lish -\frac{1}{4} ga ko'paytirishni bekor qiladi.

$$x^{2}+\frac{\frac{3}{2}}{-\frac{1}{4}}x=-\frac{4}{-\frac{1}{4}}$$

\frac{3}{2} ni -\frac{1}{4} ga bo'lish \frac{3}{2} ga k'paytirish -\frac{1}{4} ga qaytarish.

$$x^{2}-6x=-\frac{4}{-\frac{1}{4}}$$

-4 ni -\frac{1}{4} ga bo'lish -4 ga k'paytirish -\frac{1}{4} ga qaytarish.

$$x^{2}-6x=16$$

-6 ni bo‘lish, x shartining koeffitsienti, 2 ga -3 olish uchun. Keyin, -3 ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}-6x+\left(-3\right)^{2}=16+\left(-3\right)^{2}$$

-3 kvadratini chiqarish.

$$x^{2}-6x+9=16+9$$

16 ni 9 ga qo'shish.

$$x^{2}-6x+9=25$$

x^{2}-6x+9 omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x-3\right)^{2}=25$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{25}$$

Qisqartirish.

$$x-3=5 x-3=-5$$

3 ni tenglamaning ikkala tarafiga qo'shish.

$$x=8 x=-2$$