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If f(x)=x²+px+4 and g(x)=2x+1, and the minimum value of (f+g)(x) is 0, what is p?

Answer and explanation

Correct answer: p=−2±2√5

The governing concept is the minimum value of an upward-opening quadratic. Add the functions first: (f+g)(x)=x²+px+4+2x+1=x²+(p+2)x+5. Since the coefficient of x² is positive, the quadratic has a minimum. For that minimum to equal zero, the graph must touch the x-axis at its vertex, so the discriminant must be zero. Therefore (p+2)²−4(1)(5)=0, giving (p+2)²=20. Taking both square roots yields p+2=±2√5, hence p=−2±2√5. This is option A. Option B omits the shift by −2, option C gives only the shift and ignores the square-root possibilities, and option D uses an incorrect value from the discriminant equation.

Related tags

FunctionsQuadratic FunctionsMinimum ValueDiscriminantAlgebra Of Real FunctionsRelations And FunctionsMathematicsClass 11 Mcq

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