# Find the integral values of “a” for which the equation x^4-(a^2-5a+6) x^2-(a^2-3a+2)=0 has real solutions only

SAGAR SINGH - IIT DELHI
879 Points
11 years ago

Dear student,

Just put discriminant D>0

D=b2-4ac

jagdish singh singh
173 Points
11 years ago

$\textbf{Here \mathbf{x^4-(a^2-5a+6)x^2+(a^2-3a+2)=0} has real roots, Means }\\ \textbf{The Given equation has \mathbf{D\geq 0}}\\\\ \textbf{Let \mathbf{x^2=t}, where \mathbf{t\geq0} }\\\\ \mathbf{t^2-(a^2-5a+6)t+(a^2-3a+2)=0}\\\\ \textbf{Let Given equation has Roots \mathbf{t_{1}} and \mathbf{t_{2} and \mathbf{t_{1},t_{2}\geq0}}}\\\\ \textbf{So \mathbf{t_{1}+t_{2}\geq0\Leftrightarrow a^2-5a+6\geq0\Leftrightarrow (a-2).(a-3)\geq0\Leftrightarrow \boxed{a\in \left(-\infty,2\right]\cup [3,\infty)}} }\\ \textbf{And \mathbf{t_{1}.t_{2}\geq 0\Leftrightarrow a^2-3a+2\leq 0\Leftrightarrow \boxed{a\in[1,2]}}}$$\textbf{Here \mathbf{x^4-(a^2-5a+6)x^2-(a^2-3a+2)=0} has real roots, Means }\\ \textbf{Let \mathbf{x^2=t}, where \mathbf{t\geq0} }\\\\ \mathbf{t^2-(a^2-5a+6)t-(a^2-3a+2)=0}\\\\ \textbf{Let Given equation has Roots \mathbf{t_{1}} and \mathbf{t_{2} and \mathbf{t_{1},t_{2}\geq0}}}\\\\ \textbf{So \mathbf{t_{1}+t_{2}\geq0\Leftrightarrow a^2-5a+6\geq0\Leftrightarrow (a-2).(a-3)\geq0\Leftrightarrow \boxed{a\in \left(-\infty,2\right]\cup [3,\infty)}} }\\ \textbf{And \mathbf{t_{1}.t_{2}\geq 0\Leftrightarrow a^2-3a+2\leq 0\Leftrightarrow \boxed{a\in[1,2]}}}\\\\ \textbf{Now take Coommon value of \mathbf{a}, We Get Integer value \mathbf{a=1,2}}\\\\ \textbf{For Real Roots,The Given equation has \mathbf{D\geq 0}}\\\\ \textbf{At \mathbf{a=1,2, D\geq0}}\\\\ \textbf{Only Two value Which is \boxed{\boxed{\mathbf{a=1,2}}}}$