Grade 11Algebra

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

Parth Malik

15 Years agoGrade 11

2 Answers

SAGAR SINGH - IIT DELHI

15 Years ago

Dear student,

Just put discriminant D>0

D=b2-4ac

jagdish singh singh

15 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 $\boxed{\boxed{\mathbf{a=1,2}}}$}$