42x^2+13x-35=0 eching

$$42 x ^ { 2 } + 13 x - 35 = 0$$

x uchun yechish

$$x=\frac{\sqrt{6049}-13}{84}\approx 0,771134731$$

$$x=\frac{-\sqrt{6049}-13}{84}\approx -1,080658541$$

Kvadrat tenglamalaridan foydalanish qoidalari

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.

$$42x^{2}+13x-35=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 42 ni a, 13 ni b va -35 ni c bilan almashtiring.

$$x=\frac{-13±\sqrt{13^{2}-4\times 42\left(-35\right)}}{2\times 42}$$

13 kvadratini chiqarish.

$$x=\frac{-13±\sqrt{169-4\times 42\left(-35\right)}}{2\times 42}$$

-4 ni 42 marotabaga ko'paytirish.

$$x=\frac{-13±\sqrt{169-168\left(-35\right)}}{2\times 42}$$

-168 ni -35 marotabaga ko'paytirish.

$$x=\frac{-13±\sqrt{169+5880}}{2\times 42}$$

169 ni 5880 ga qo'shish.

$$x=\frac{-13±\sqrt{6049}}{2\times 42}$$

2 ni 42 marotabaga ko'paytirish.

$$x=\frac{-13±\sqrt{6049}}{84}$$

x=\frac{-13±\sqrt{6049}}{84} tenglamasini yeching, bunda ± musbat. -13 ni \sqrt{6049} ga qo'shish.

$$x=\frac{\sqrt{6049}-13}{84}$$

x=\frac{-13±\sqrt{6049}}{84} tenglamasini yeching, bunda ± manfiy. -13 dan \sqrt{6049} ni ayirish.

$$x=\frac{-\sqrt{6049}-13}{84}$$

Tenglama yechildi.

$$x=\frac{\sqrt{6049}-13}{84} x=\frac{-\sqrt{6049}-13}{84}$$

Kvadratni yakunlash bo'yicha qadamlar

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

$$42x^{2}+13x-35=0$$

35 ni tenglamaning ikkala tarafiga qo'shish.

$$42x^{2}+13x-35-\left(-35\right)=-\left(-35\right)$$

O‘zidan -35 ayirilsa 0 qoladi.

$$42x^{2}+13x=-\left(-35\right)$$

0 dan -35 ni ayirish.

$$42x^{2}+13x=35$$

Ikki tarafini 42 ga bo‘ling.

$$\frac{42x^{2}+13x}{42}=\frac{35}{42}$$

42 ga bo'lish 42 ga ko'paytirishni bekor qiladi.

$$x^{2}+\frac{13}{42}x=\frac{35}{42}$$

\frac{35}{42} ulushini 7 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x^{2}+\frac{13}{42}x=\frac{5}{6}$$

\frac{13}{42} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{13}{84} olish uchun. Keyin, \frac{13}{84} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+\frac{13}{42}x+\left(\frac{13}{84}\right)^{2}=\frac{5}{6}+\left(\frac{13}{84}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{13}{84} kvadratini chiqarish.

$$x^{2}+\frac{13}{42}x+\frac{169}{7056}=\frac{5}{6}+\frac{169}{7056}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{5}{6} ni \frac{169}{7056} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}+\frac{13}{42}x+\frac{169}{7056}=\frac{6049}{7056}$$

x^{2}+\frac{13}{42}x+\frac{169}{7056} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+\frac{13}{84}\right)^{2}=\frac{6049}{7056}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{13}{84}\right)^{2}}=\sqrt{\frac{6049}{7056}}$$

Qisqartirish.

$$x+\frac{13}{84}=\frac{\sqrt{6049}}{84} x+\frac{13}{84}=-\frac{\sqrt{6049}}{84}$$

Tenglamaning ikkala tarafidan \frac{13}{84} ni ayirish.

$$x=\frac{\sqrt{6049}-13}{84} x=\frac{-\sqrt{6049}-13}{84}$$