2.6-bo'lim : Kvadrat tenglamalar - II qism
12. Quyidagi kvadrat tenglamani yechish uchun kvadrat formuladan foydalaning.
\[169 - 20t + 4{t^2} = 0\]Yechim
Birinchidan, kvadrat tenglamani standart shaklda olishimiz kerak. Bu,
\[4{t^2} - 20t + 169 = 0\]E'tibor bering, bu holda biz atamalarni kamayuvchi ko'rsatkichlarga ega bo'lish uchun o'zgartirdik.
2-qadam
Endi kvadrat formula uchun qiymatlarni aniqlashimiz kerak.
\[a = 4\hspace{0.25in}b = - 20\hspace{0.25in}c = 169\]3-qadam
Bularni kvadratik formulaga kiritish,
\[t = \frac{{ - \left( { - 20} \right) \pm \sqrt {{{\left( { - 20} \right)}^2} - 4\left( 4 \right)\left( {169} \right)} }}{{2\left( 4 \right)}} = \frac{{20 \pm \sqrt { - 2304} }}{8} = \frac{{20 \pm 48i}}{8} = \frac{5}{2} \pm 6i\]Bu tenglamaning ikkita yechimi quyidagicha bo'ladi: \[\require{bbox} \bbox[2pt,border:1px solid black]{{t = \frac{5}{2} - 6i\,\,{\mbox{and }}t = \frac{5}{2} + 6i}}\].