Question:
The equation \(\sqrt{3x^2+x+5}=x-3\), where \(x\) is real, has :
The equation \(\sqrt{3x^2+x+5}=x-3\), where \(x\) is real, has :
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Square both sides, then check the roots against the condition \(x-3\ge 0\).
Updated On: Nov 11, 2026
- no solution
- exactly one solution
- exactly two solutions
- exactly four solutions.
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The Correct Option is A
Solution and Explanation
The left side is non-negative, so we need \(x-3\ge 0\), i.e. \(x\ge 3\).
Squaring: \(3x^2+x+5=x^2-6x+9\Rightarrow 2x^2+7x-4=0\Rightarrow (2x-1)(x+4)=0\).
So \(x=\frac12\) or \(x=-4\). Both are less than 3, hence both are extraneous (they give \(x-3<0\)).
Answer: (1) no solution
Squaring: \(3x^2+x+5=x^2-6x+9\Rightarrow 2x^2+7x-4=0\Rightarrow (2x-1)(x+4)=0\).
So \(x=\frac12\) or \(x=-4\). Both are less than 3, hence both are extraneous (they give \(x-3<0\)).
Answer: (1) no solution
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