Профиль №18. Исследование дискриминанта и применение теоремы Виета. Задача 27В

ОТВЕТ:  \(-\frac{3}{{10}}.\)

\(\left\{ {\begin{array}{*{20}{c}}{x-3y = -1,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{{x^2} + 6xy + 9{y^2}-10\,a\,x-30\,a\,y + 125{a^2} + 60a + 9 = 0}\end{array}} \right.\;\;\;\; \Leftrightarrow \)

\( \Leftrightarrow \;\;\;\;\left\{ {\begin{array}{*{20}{c}}{x-3y = -1,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{{{\left( {x + 3y} \right)}^2}-10\,a\,\left( {x + 3y} \right) + 125{a^2} + 60a + 9 = 0.}\end{array}} \right.\)

Пусть  \(x + 3y = t.\)  Тогда полученная система примет вид:

\(\left\{ {\begin{array}{*{20}{c}}{x-3y = -1,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{{t^2}-10\,a\,t + 125{a^2} + 60a + 9 = 0.}\end{array}} \right.\)

При этом система  \(\left\{ {\begin{array}{*{20}{c}}{x-3y = -1,}\\{x + 3y = t\;\;\;}\end{array}} \right.\)  будет иметь одно решение при \(t \in R,\)  так как  \(\frac{1}{1} \ne \frac{{-3}}{3}.\)

Следовательно, исходная система будет иметь одно решение, если уравнение  \({t^2}-10\,a\,t + 125{a^2} + 60a + 9 = 0\)  будет иметь один корень, а это возможно только в случае, если  \(D = 0:\)

\(D = 100{a^2}-500{a^2}-240a-36 = 0\;\;\;\; \Leftrightarrow \;\;\;\;-400{a^2}-240a-36 = 0\;\;\;\; \Leftrightarrow \)

\( \Leftrightarrow \;\;\;\;100{a^2} + 60a + 9 = 0\;\;\;\; \Leftrightarrow \;\;\;\;{\left( {10a + 3} \right)^2} = 0\;\;\;\; \Leftrightarrow \;\;\;\;a = -\frac{3}{{10}}.\)

Ответ:  \(-\frac{3}{{10}}.\)