گام به گام تمرین صفحه 31 درس 1 ریاضی و آمار (1) (معادلۀ درجۀ دوم)

\(\begin{array}{l} \Rightarrow \left\{ \begin{array}{l}a = 1\\\\b = - 1\\\\c = 0\end{array} \right. \Rightarrow \Delta = {b^2} - 4ac = {( - 1)^2} - 4(1)(0) = 1\\\\ \Rightarrow x = \frac{{ - b \pm \sqrt \Delta }}{{2a}} = \frac{{1 \pm \sqrt 1 }}{{2(1)}} = \frac{{1 \pm 1}}{2} \Rightarrow \left\{ \begin{array}{l}x = \frac{{1 - 1}}{2} = 0\\\\x = \frac{{1 + 1}}{2} = 1\end{array} \right.\end{array}\)

\(\begin{array}{l} \Rightarrow \left\{ \begin{array}{l}a = 2\\\\b = 1\\\\c = - 1\end{array} \right. \Rightarrow \Delta = {b^2} - 4ac = {1^2} - 4(2)( - 1) = 1 + 8 = 9\\\\ \Rightarrow x = \frac{{ - b \pm \sqrt \Delta }}{{2a}} = \frac{{ - 1 \pm \sqrt 9 }}{{2(2)}} = \frac{{ - 1 \pm 3}}{4}\\\\ \Rightarrow \left\{ \begin{array}{l}x = \frac{{ - 1 - 3}}{4} = - 1\\\\x = \frac{{ - 1 + 3}}{4} = \frac{1}{2}\end{array} \right.\end{array}\)

\(\begin{array}{l} \Rightarrow \left\{ \begin{array}{l}a = 4\\\\b = - 4\\\\c = 1\end{array} \right. \Rightarrow \Delta = {b^2} - 4ac = {( - 4)^2} - 4(4)(1) = 0\\\\ \Rightarrow x = \frac{{ - b \pm \sqrt \Delta }}{{2a}} = \frac{{4 \pm \sqrt 0 }}{{2(4)}} = \frac{4}{8} = \frac{1}{2}\end{array}\)

\(\begin{array}{l} \Rightarrow \left\{ \begin{array}{l}a = 1\\\\b = 17\\\\c = - 18\end{array} \right. \Rightarrow \Delta = {b^2} - 4ac = {(17)^2} - 4(1)( - 18) = 361\\\\ \Rightarrow x = \frac{{ - b \pm \sqrt \Delta }}{{2a}} = \frac{{ - 17 \pm \sqrt {361} }}{{2(1)}} = \frac{{ - 17 \pm 19}}{2}\\\\ \Rightarrow \left\{ \begin{array}{l}x = \frac{{ - 17 - 19}}{2} = - 18\\\\x = \frac{{ - 17 + 19}}{2} = 1\end{array} \right.\end{array}\)

\(\begin{array}{l} \Rightarrow \left\{ \begin{array}{l}a = 3\\\\b = - 1\\\\c = 4\end{array} \right. \Rightarrow \Delta = {b^2} - 4ac = {( - 1)^2} - 4(3)(4) = - 41\\\\ \Rightarrow \Delta < 0\end{array}\)

به دلیل \(\Delta < 0\)، معادله فوق جواب ندارد.

\(\begin{array}{l} \Rightarrow \left\{ \begin{array}{l}a = 1\\\\b = \sqrt 3 \\\\c = - 1\end{array} \right. \Rightarrow \Delta = {b^2} - 4ac = {(\sqrt 3 )^2} - 4(1)( - 1) = 7\\\\ \Rightarrow x = \frac{{ - b \pm \sqrt \Delta }}{{2a}} = \frac{{ - \sqrt 3 \pm \sqrt 7 }}{{2(1)}} = \frac{{ - \sqrt 3 \pm \sqrt 7 }}{2}\\\\ \Rightarrow \left\{ \begin{array}{l}x = \frac{{ - \sqrt 3 - \sqrt 7 }}{2}\\\\x = \frac{{ - \sqrt 3 + \sqrt 7 }}{2}\end{array} \right.\end{array}\)

\(\begin{array}{l} \Rightarrow \left\{ \begin{array}{l}a = 2\\\\b = - 3\\\\c = - 5\end{array} \right. \Rightarrow \Delta = {b^2} - 4ac = {( - 3)^2} - 4(2)( - 5) = 49\\\\ \Rightarrow x = \frac{{ - b \pm \sqrt \Delta }}{{2a}} = \frac{{ - ( - 3) \pm \sqrt {49} }}{{2(2)}} = \frac{{3 \pm 7}}{4}\\\\ \Rightarrow \left\{ \begin{array}{l}{x_1} = \frac{{3 - 7}}{4} = - 1\\\\{x_2} = \frac{{3 + 7}}{4} = \frac{5}{2}\end{array} \right. \Rightarrow {x_1} \times {x_2} = ( - 1) \times \frac{5}{2} = - \frac{5}{2}\end{array}\)

روش اول:

\(\begin{array}{l}x = 4\,\,\,\,\,(1)\\\\\Delta = {( - a)^2} - 4(2)(28) = {a^2} - 224\,\,\,\,\,(2)\\\\x = \frac{{ - ( - a) \pm \sqrt \Delta }}{{2(2)}} = \frac{{a \pm \sqrt \Delta }}{4}\\\\ \Rightarrow 4x = a \pm \sqrt \Delta \Rightarrow \pm \sqrt \Delta = 4x - a\mathop \Rightarrow \limits^{{{()}^2}} \Delta = {(4x - a)^2}\\\\(1)\,\,\,,\,\,\,(2) \Rightarrow {a^2} - 224 = {(16 - a)^2}\\\\ \Rightarrow {a^2} - 224 = 256 - 32a + {a^2} \Rightarrow 32a = 480 \Rightarrow a = 15\\\\ \Rightarrow \Delta = {a^2} - 224 = {(15)^2} - 224 = 225 - 224 = 1\\\\ \Rightarrow x = \frac{{a \pm \sqrt \Delta }}{4} = \frac{{15 \pm \sqrt 1 }}{4} = \frac{{15 \pm 1}}{4} \Rightarrow \left\{ \begin{array}{l}{x_1} = \frac{{15 + 1}}{4} = 4\\\\{x_2} = \frac{{15 - 1}}{4} = \frac{7}{2}\end{array} \right.\end{array}\)

روش دوم:

یک روش دیگر برای محاسبه دومین ریشه وجود دارد و آن به صورت زیر آمده است:

برای معادلات درجه 2 به فرم \(a{x^2} + bx + c = 0\) ، رابطه زیر بین حاصل ضرب دو ریشه ی ضرایب معادله برقرار است:

\({x_1} \times {x_2} = \frac{c}{a}\)

بنابراین برای معادله \(2{x^2} - ax + 28 = 0\) داریم:

\(\left. \begin{array}{l}{x_1} \times {x_2} = \frac{{28}}{2} = 14\\\\{x_1} = 4\end{array} \right\} \Rightarrow 4 \times {x_2} = 14 \Rightarrow {x_2} = \frac{{14}}{4} = \frac{7}{2}\)

\( = \frac{{2x \times (3x + 6)}}{2} = 3{x^2} + 6x\) مساحت مثلث

\( = (3x + 2) \times (x + 1) = 3{x^2} + 5x + 2\) مساحت مستطیل

\( \Rightarrow 3{x^2} + 6x = 3{x^2} + 5x + 2 \Rightarrow x = 2\) مساحت مستطیل = مساحت مثلث

\( = 3x + 2 = 8\) طول مستطیل

\( = x + 1 = 3\) عرض مستطیل

\(\Delta = {a^2} - 4(1)( - 1) = {a^2} + 4 > 0\)

این مقدار همیشه مثبت است زیرا به ازای همه مقادیر a معادله مثبت بوده در نتیجه دارای دو ریشه است.

\(\Delta = {( - 1)^2} - 4(1)(a) = 1 - 4a \ge 0 \Rightarrow 4a \le 1 \Rightarrow a \le \frac{1}{4}\)

این معادله به ازای مقادیر a کوچکتر از \(\frac{1}{4}\) و مساوی آن مثبت بوده، در نتیجه در این بازه دارای دو ریشه است.

\(\begin{array}{l}a{x^2} + bx + c = 0 \Rightarrow \Delta = {b^2} - 4ac = {(a + c)^2} - 4ac\\\\ \Rightarrow \Delta = {a^2} + 2ac + {c^2} - 4ac = {a^2} - 2ac + {c^2} = {(a - c)^2} > 0\\\\ \Rightarrow x = \frac{{ - b \pm \sqrt \Delta }}{{2a}} = \frac{{ - (a + c) \pm \sqrt {{{(a - c)}^2}} }}{{2a}}\\\\ \Rightarrow x = \frac{{ - a - c \pm (a - c)}}{{2a}}\\\\ \Rightarrow \left\{ \begin{array}{l}{x_1} = \frac{{ - a - c - (a - c)}}{{2a}} = \frac{{ - 2a}}{{2a}} = - 1\\\\{x_2} = \frac{{ - a - c + (a - c)}}{{2a}} = \frac{{ - 2c}}{{2a}} = - \frac{c}{a}\end{array} \right.\end{array}\)

\(\begin{array}{l}x = \frac{{ - b \pm \sqrt \Delta }}{{2a}} \Rightarrow \left\{ \begin{array}{l}{x_1} = \frac{{ - b - \sqrt \Delta }}{{2a}}\\\\{x_2} = \frac{{ - b + \sqrt \Delta }}{{2a}}\end{array} \right.\\\\ \Rightarrow {x_1} \times {x_2} = \frac{{ - b - \sqrt \Delta }}{{2a}} \times \frac{{ - b + \sqrt \Delta }}{{2a}} = \frac{{{{( - b)}^2} - {{(\sqrt \Delta )}^2}}}{{4{a^2}}} = \\\\\frac{{{b^2} - \Delta }}{{4{a^2}}} = \frac{{{b^2} - ({b^2} - 4ac)}}{{4{a^2}}} = \frac{{4ac}}{{4{a^2}}} = \frac{c}{a}\end{array}\)

\(\begin{array}{l}a + b + c = 0 \Rightarrow b = - (a + c)\\\\a{x^2} + bx + c = 0 \Rightarrow \Delta = {b^2} - 4ac = {( - (a + c))^2} - 4ac\\\\ \Rightarrow \Delta = {a^2} + 2ac + {c^2} - 4ac = {a^2} - 2ac + {c^2} = {(a - c)^2} > 0\\\\ \Rightarrow x = \frac{{ - b \pm \sqrt \Delta }}{{2a}} = \frac{{ - ( - (a + c)) \pm \sqrt {{{(a - c)}^2}} }}{{2a}}\\\\ \Rightarrow x = \frac{{a + c \pm (a - c)}}{{2a}}\\\\ \Rightarrow \left\{ \begin{array}{l}{x_1} = \frac{{a + c - (a - c)}}{{2a}} = \frac{{2c}}{{2a}} = \frac{c}{a}\\\\{x_2} = \frac{{a + c + (a - c)}}{{2a}} = \frac{{2a}}{{2a}} = 1\end{array} \right.\end{array}\)