(x-35)*(x+38-25)=253575 eching

$$(x-35) \times (x+38-25)=253575$$

x uchun yechish

$$x=3\sqrt{28239}+11\approx 515,133910782$$

$$x=11-3\sqrt{28239}\approx -493,133910782$$

Kvadrat tenglamalaridan foydalanish qoidalari

13 olish uchun 38 dan 25 ni ayirish.

$$\left(x-35\right)\left(x+13\right)=253575$$

x-35 ga x+13 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

$$x^{2}-22x-455=253575$$

Ikkala tarafdan 253575 ni ayirish.

$$x^{2}-22x-455-253575=0$$

-254030 olish uchun -455 dan 253575 ni ayirish.

$$x^{2}-22x-254030=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 1 ni a, -22 ni b va -254030 ni c bilan almashtiring.

$$x=\frac{-\left(-22\right)±\sqrt{\left(-22\right)^{2}-4\left(-254030\right)}}{2}$$

-22 kvadratini chiqarish.

$$x=\frac{-\left(-22\right)±\sqrt{484-4\left(-254030\right)}}{2}$$

-4 ni -254030 marotabaga ko'paytirish.

$$x=\frac{-\left(-22\right)±\sqrt{484+1016120}}{2}$$

484 ni 1016120 ga qo'shish.

$$x=\frac{-\left(-22\right)±\sqrt{1016604}}{2}$$

1016604 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-22\right)±6\sqrt{28239}}{2}$$

-22 ning teskarisi 22 ga teng.

$$x=\frac{22±6\sqrt{28239}}{2}$$

x=\frac{22±6\sqrt{28239}}{2} tenglamasini yeching, bunda ± musbat. 22 ni 6\sqrt{28239} ga qo'shish.

$$x=\frac{6\sqrt{28239}+22}{2}$$

22+6\sqrt{28239} ni 2 ga bo'lish.

$$x=3\sqrt{28239}+11$$

x=\frac{22±6\sqrt{28239}}{2} tenglamasini yeching, bunda ± manfiy. 22 dan 6\sqrt{28239} ni ayirish.

$$x=\frac{22-6\sqrt{28239}}{2}$$

22-6\sqrt{28239} ni 2 ga bo'lish.

$$x=11-3\sqrt{28239}$$

Tenglama yechildi.

$$x=3\sqrt{28239}+11 x=11-3\sqrt{28239}$$

Kvadratni yakunlash bo'yicha qadamlar

13 olish uchun 38 dan 25 ni ayirish.

$$\left(x-35\right)\left(x+13\right)=253575$$

x-35 ga x+13 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

$$x^{2}-22x-455=253575$$

455 ni ikki tarafga qo’shing.

$$x^{2}-22x=253575+455$$

254030 olish uchun 253575 va 455'ni qo'shing.

$$x^{2}-22x=254030$$

-22 ni bo‘lish, x shartining koeffitsienti, 2 ga -11 olish uchun. Keyin, -11 ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}-22x+\left(-11\right)^{2}=254030+\left(-11\right)^{2}$$

-11 kvadratini chiqarish.

$$x^{2}-22x+121=254030+121$$

254030 ni 121 ga qo'shish.

$$x^{2}-22x+121=254151$$

x^{2}-22x+121 omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x-11\right)^{2}=254151$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-11\right)^{2}}=\sqrt{254151}$$

Qisqartirish.

$$x-11=3\sqrt{28239} x-11=-3\sqrt{28239}$$

11 ni tenglamaning ikkala tarafiga qo'shish.

$$x=3\sqrt{28239}+11 x=11-3\sqrt{28239}$$