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If \(p(x)=3x^2-12x+15\), what is the correct conclusion about its real zeroes?

Answer and explanation

Correct answer: No real zeroes

Here, \(a=3\), \(b=-12\), and \(c=15\). The discriminant is \(D=b^2-4ac=(-12)^2-4(3)(15)=-36<0\), so the quadratic polynomial has no real zeroes. Equivalently, \(p(x)=3[(x-2)^2+1]\), which is positive for every real \(x\). Thus option B is incorrect because equal real zeroes require \(D=0\). Exam tip: for a quadratic polynomial, \(D<0\) means that it has no real zeroes.

Related tags

Real ZeroesQuadratic PolynomialDiscriminantCompleting The SquarePolynomials In One VariablePolynomialsMathematicsClass 10 Mcq

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