t^2+11+t-42=0 eching

$${ t }^{ 2 } +11+t-42 = 0$$

t uchun yechish

$$t = \frac{5 \sqrt{5} - 1}{2} \approx 5,090169944$$

$$t=\frac{-5\sqrt{5}-1}{2}\approx -6,090169944$$

Kvadrat tenglamalaridan foydalanish qoidalari

-31 olish uchun 11 dan 42 ni ayirish.

$$t^{2}-31+t=0$$

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.

$$t^{2}+t-31=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 1 ni a, 1 ni b va -31 ni c bilan almashtiring.

$$t=\frac{-1±\sqrt{1^{2}-4\left(-31\right)}}{2}$$

1 kvadratini chiqarish.

$$t=\frac{-1±\sqrt{1-4\left(-31\right)}}{2}$$

-4 ni -31 marotabaga ko'paytirish.

$$t=\frac{-1±\sqrt{1+124}}{2}$$

1 ni 124 ga qo'shish.

$$t=\frac{-1±\sqrt{125}}{2}$$

125 ning kvadrat ildizini chiqarish.

$$t=\frac{-1±5\sqrt{5}}{2}$$

t=\frac{-1±5\sqrt{5}}{2} tenglamasini yeching, bunda ± musbat. -1 ni 5\sqrt{5} ga qo'shish.

$$t=\frac{5\sqrt{5}-1}{2}$$

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

$$t=\frac{-5\sqrt{5}-1}{2}$$

Tenglama yechildi.

$$t=\frac{5\sqrt{5}-1}{2} t=\frac{-5\sqrt{5}-1}{2}$$

Kvadratni yakunlash bo'yicha qadamlar

-31 olish uchun 11 dan 42 ni ayirish.

$$t^{2}-31+t=0$$

31 ni ikki tarafga qo’shing. Har qanday songa nolni qo‘shsangiz, o‘zi chiqadi.

$$t^{2}+t=31$$

1 ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{1}{2} olish uchun. Keyin, \frac{1}{2} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$t^{2}+t+\left(\frac{1}{2}\right)^{2}=31+\left(\frac{1}{2}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{1}{2} kvadratini chiqarish.

$$t^{2}+t+\frac{1}{4}=31+\frac{1}{4}$$

31 ni \frac{1}{4} ga qo'shish.

$$t^{2}+t+\frac{1}{4}=\frac{125}{4}$$

t^{2}+t+\frac{1}{4} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(t+\frac{1}{2}\right)^{2}=\frac{125}{4}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(t+\frac{1}{2}\right)^{2}}=\sqrt{\frac{125}{4}}$$

Qisqartirish.

$$t+\frac{1}{2}=\frac{5\sqrt{5}}{2} t+\frac{1}{2}=-\frac{5\sqrt{5}}{2}$$

Tenglamaning ikkala tarafidan \frac{1}{2} ni ayirish.

$$t=\frac{5\sqrt{5}-1}{2} t=\frac{-5\sqrt{5}-1}{2}$$