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Question
The number of real roots of the equation \[( x^2 + 2x )^2 - (x + 1 )^2 - 55 = 0\] is
Options
2
1
4
none of these
Solution
2
Explanation:
\[\left( x^2 + 2x \right)^2 - \left( x + 1 \right)^2 - 55 = 0\]
\[ \Rightarrow \left( x^2 + 2x + 1 - 1 \right)^2 - \left( x + 1 \right)^2 - 55 = 0\]
\[ \Rightarrow \left\{ \left( x + 1 \right)^2 - 1 \right\}^2 - \left( x + 1 \right)^2 - 55 = 0\]
\[ \Rightarrow \left\{ \left( x + 1 \right)^2 \right\}^2 + 1 - 3 \left( x + 1 \right)^2 - 55 = 0\]
\[ \Rightarrow \left\{ \left( x + 1 \right)^2 \right\}^2 - 3 \left( x + 1 \right)^2 - 54 = 0\]
\[\text { Let } p = \left( x + 1 \right)^2 \]
\[ \Rightarrow p^2 - 3p - 54 = 0\]
\[ \Rightarrow p^2 - 9p + 6p - 54 = 0\]
\[ \Rightarrow \left( p + 6 \right)\left( p - 9 \right) = 0\]
\[ \Rightarrow p = 9 \text { or }p = - 6\]
\[\text { Rejecting }p = - 6\]
\[ \Rightarrow \left( x + 1 \right)^2 = 9\]
\[ \Rightarrow x^2 + 2x - 8 = 0\]
\[ \Rightarrow x^2 + 4x - 2x - 8 = 0\]
\[ \Rightarrow \left( x + 4 \right)\left( x - 2 \right) = 0\]
\[ \Rightarrow x = 2, x = - 4\]
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