11seeing the domain of x ;
x lies in between (9-a) and (9+a)
and between -4 and 4.
by solving the equation ,
a2+2ax-7 >0
for this discriminant must be negetve, i.e,
x2+7 < 0
which is nt possible with real valued x :(
where am i going wrong??
$\textbf{Find no. of integral values of $\mathbf{a}$ such that\\\\ $\mathbf{\sqrt{9-a^2+2ax-x^2}>\sqrt{16-x^2}$}}$\\\\ $\textbf{for atleast one positive Real value of $\mathbf{x}$}.$
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4 Answers
1111oh.. i have commited a mistake there..
after simplifying, the equation comes out to be,
a2 - 2ax + 7 <0
for this discriminant should be negetve.. thus,
x2 - 7 < 0
x lies between √7 and - √7
how to proceed after that...
