The smallest positive integral value of a, for which all the roots of x^4-a x^2+9=0 are re…

Question

The smallest positive integral value of $a$, for which all the roots of $x^4-a x^2+9=0$ are real and distinct, is equal to

Accepted Answer

Given equation

${x}^{4}-a{x}^{2}+9=0---$

let ${x}^{2}=t$, equation (1) become

${t}^{2}-at+9=0---$

If ${t}_{1}$ \& ${t}_{2}$ are roots of equation (2) then four distinct roots of equation (1) is ${-\sqrt{{t}_{1}},-\sqrt{{t}_{2}},\sqrt{{t}_{1}},\sqrt{{t}_{2}}}$

roots of this equation are real and distinct if $D>0$

$\begin{gathered} (-a{)}^{2}-4	imes 9>0 \ \Rightarrow {a}^{2}-36>0 \end{gathered}$

so $a\in (-\infty,-6)\cup (6,\infty)$

so smallest positive integral value of $a$ is 7 as $a>6$.

Solutions · 1