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For which interval of the parameter \(a\) does the quadratic equation \(3x^2-2(2a+1)x+(a^2+a+1)=0\) have no real roots?

Answer and explanation

Correct answer: \(-2<a<1\)

A quadratic equation has no real roots when its discriminant satisfies \(D<0\). Here, \(D=[-2(2a+1)]^2-4(3)(a^2+a+1)=4(a^2+a-2)=4(a+2)(a-1)\). Therefore, \((a+2)(a-1)<0\), which gives \(-2<a<1\). At the endpoints \(a=-2\) and \(a=1\), \(D=0\), so they are not included. Exam tip: \(D<0\) indicates no real roots, whereas \(D=0\) indicates equal real roots.

Related tags

Quadratic EquationsNature Of RootsDiscriminantNo Real RootsInequalitiesMathematicsClass 10 Mcq

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