2=-1/4x^2+5/2x eching

$$2 = - \frac { 1 } { 4 } x ^ { 2 } + \frac { 5 } { 2 } x$$

x uchun yechish

$$x=\sqrt{17}+5\approx 9,123105626$$

$$x=5-\sqrt{17}\approx 0,876894374$$

Kvadrat tenglamalaridan foydalanish qoidalari

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$-\frac{1}{4}x^{2}+\frac{5}{2}x=2$$

Ikkala tarafdan 2 ni ayirish.

$$-\frac{1}{4}x^{2}+\frac{5}{2}x-2=0$$

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{5}{2} ni b va -2 ni c bilan almashtiring.

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

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{5}{2} kvadratini chiqarish.

$$x=\frac{-\frac{5}{2}±\sqrt{\frac{25}{4}-4\left(-\frac{1}{4}\right)\left(-2\right)}}{2\left(-\frac{1}{4}\right)}$$

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

$$x=\frac{-\frac{5}{2}±\sqrt{\frac{25}{4}-2}}{2\left(-\frac{1}{4}\right)}$$

\frac{25}{4} ni -2 ga qo'shish.

$$x=\frac{-\frac{5}{2}±\sqrt{\frac{17}{4}}}{2\left(-\frac{1}{4}\right)}$$

\frac{17}{4} ning kvadrat ildizini chiqarish.

$$x=\frac{-\frac{5}{2}±\frac{\sqrt{17}}{2}}{2\left(-\frac{1}{4}\right)}$$

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

$$x=\frac{-\frac{5}{2}±\frac{\sqrt{17}}{2}}{-\frac{1}{2}}$$

x=\frac{-\frac{5}{2}±\frac{\sqrt{17}}{2}}{-\frac{1}{2}} tenglamasini yeching, bunda ± musbat. -\frac{5}{2} ni \frac{\sqrt{17}}{2} ga qo'shish.

$$x=\frac{\sqrt{17}-5}{-\frac{1}{2}\times 2}$$

\frac{-5+\sqrt{17}}{2} ni -\frac{1}{2} ga bo'lish \frac{-5+\sqrt{17}}{2} ga k'paytirish -\frac{1}{2} ga qaytarish.

$$x=5-\sqrt{17}$$

x=\frac{-\frac{5}{2}±\frac{\sqrt{17}}{2}}{-\frac{1}{2}} tenglamasini yeching, bunda ± manfiy. -\frac{5}{2} dan \frac{\sqrt{17}}{2} ni ayirish.

$$x=\frac{-\sqrt{17}-5}{-\frac{1}{2}\times 2}$$

\frac{-5-\sqrt{17}}{2} ni -\frac{1}{2} ga bo'lish \frac{-5-\sqrt{17}}{2} ga k'paytirish -\frac{1}{2} ga qaytarish.

$$x=\sqrt{17}+5$$

Tenglama yechildi.

$$x=5-\sqrt{17} x=\sqrt{17}+5$$

Kvadratni yakunlash bo'yicha qadamlar

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$-\frac{1}{4}x^{2}+\frac{5}{2}x=2$$

Ikkala tarafini -4 ga ko‘paytiring.

$$\frac{-\frac{1}{4}x^{2}+\frac{5}{2}x}{-\frac{1}{4}}=\frac{2}{-\frac{1}{4}}$$

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

$$x^{2}+\frac{\frac{5}{2}}{-\frac{1}{4}}x=\frac{2}{-\frac{1}{4}}$$

\frac{5}{2} ni -\frac{1}{4} ga bo'lish \frac{5}{2} ga k'paytirish -\frac{1}{4} ga qaytarish.

$$x^{2}-10x=\frac{2}{-\frac{1}{4}}$$

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

$$x^{2}-10x=-8$$

-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.

$$x^{2}-10x+\left(-5\right)^{2}=-8+\left(-5\right)^{2}$$

-5 kvadratini chiqarish.

$$x^{2}-10x+25=-8+25$$

-8 ni 25 ga qo'shish.

$$x^{2}-10x+25=17$$

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.

$$\left(x-5\right)^{2}=17$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-5\right)^{2}}=\sqrt{17}$$

Qisqartirish.

$$x-5=\sqrt{17} x-5=-\sqrt{17}$$

5 ni tenglamaning ikkala tarafiga qo'shish.

$$x=\sqrt{17}+5 x=5-\sqrt{17}$$