65=2*x^2+9*x+5= eching

$$65=2 \times { x }^{ 2 } +9 \times x+5=$$

x uchun yechish

$$x = \frac{\sqrt{561} - 9}{4} \approx 3,671359641$$

$$x=\frac{-\sqrt{561}-9}{4}\approx -8,171359641$$

Kvadrat tenglamalaridan foydalanish qoidalari

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$2x^{2}+9x+5=65$$

Ikkala tarafdan 65 ni ayirish.

$$2x^{2}+9x+5-65=0$$

-60 olish uchun 5 dan 65 ni ayirish.

$$2x^{2}+9x-60=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 2 ni a, 9 ni b va -60 ni c bilan almashtiring.

$$x=\frac{-9±\sqrt{9^{2}-4\times 2\left(-60\right)}}{2\times 2}$$

9 kvadratini chiqarish.

$$x=\frac{-9±\sqrt{81-4\times 2\left(-60\right)}}{2\times 2}$$

-4 ni 2 marotabaga ko'paytirish.

$$x=\frac{-9±\sqrt{81-8\left(-60\right)}}{2\times 2}$$

-8 ni -60 marotabaga ko'paytirish.

$$x=\frac{-9±\sqrt{81+480}}{2\times 2}$$

81 ni 480 ga qo'shish.

$$x=\frac{-9±\sqrt{561}}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$x=\frac{-9±\sqrt{561}}{4}$$

x=\frac{-9±\sqrt{561}}{4} tenglamasini yeching, bunda ± musbat. -9 ni \sqrt{561} ga qo'shish.

$$x=\frac{\sqrt{561}-9}{4}$$

x=\frac{-9±\sqrt{561}}{4} tenglamasini yeching, bunda ± manfiy. -9 dan \sqrt{561} ni ayirish.

$$x=\frac{-\sqrt{561}-9}{4}$$

Tenglama yechildi.

$$x=\frac{\sqrt{561}-9}{4} x=\frac{-\sqrt{561}-9}{4}$$

Kvadratni yakunlash bo'yicha qadamlar

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$2x^{2}+9x+5=65$$

Ikkala tarafdan 5 ni ayirish.

$$2x^{2}+9x=65-5$$

60 olish uchun 65 dan 5 ni ayirish.

$$2x^{2}+9x=60$$

Ikki tarafini 2 ga bo‘ling.

$$\frac{2x^{2}+9x}{2}=\frac{60}{2}$$

2 ga bo'lish 2 ga ko'paytirishni bekor qiladi.

$$x^{2}+\frac{9}{2}x=\frac{60}{2}$$

60 ni 2 ga bo'lish.

$$x^{2}+\frac{9}{2}x=30$$

\frac{9}{2} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{9}{4} olish uchun. Keyin, \frac{9}{4} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+\frac{9}{2}x+\left(\frac{9}{4}\right)^{2}=30+\left(\frac{9}{4}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{9}{4} kvadratini chiqarish.

$$x^{2}+\frac{9}{2}x+\frac{81}{16}=30+\frac{81}{16}$$

30 ni \frac{81}{16} ga qo'shish.

$$x^{2}+\frac{9}{2}x+\frac{81}{16}=\frac{561}{16}$$

x^{2}+\frac{9}{2}x+\frac{81}{16} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+\frac{9}{4}\right)^{2}=\frac{561}{16}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{9}{4}\right)^{2}}=\sqrt{\frac{561}{16}}$$

Qisqartirish.

$$x+\frac{9}{4}=\frac{\sqrt{561}}{4} x+\frac{9}{4}=-\frac{\sqrt{561}}{4}$$

Tenglamaning ikkala tarafidan \frac{9}{4} ni ayirish.

$$x=\frac{\sqrt{561}-9}{4} x=\frac{-\sqrt{561}-9}{4}$$