Virtual Math Tutor: Math problem #48

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Sunday, December 19, 2010

Math problem #48

Math problem: It is known that and    are the roots of the quadratic equation


Derive a quadratic equation whose roots are   and .

Discussion and hints: Let us consider a general quadratic equation


Without loss of generality we may assume that To derive the formulas for the roots of the general quadratic equation (1) we complete the square on the left hand side to get





Introduce the discriminant of (1). It is clear from the formula above that if the quadratic equation (1) has two real roots, which are given by the quadratic formula:



If then (1) admits one (double) real root



If then (1) has two complex-conjugate roots



From the formulas that we have derived above, it is easy to compute (Vi├Ęte's formulas)






Figure 1 


Let us also say a few words about maximum and minimum of the quadratic polynomial function


If then the polynomial has no maximum value, but has the minimum of at If then the polynomial ha no minimum, but it has the maximum value of at

The graph of the function is a parabola. Its vertex is at the point



If the parabola is concave up, if the parabola is concave down. The vertical axis of such a parabola is the line



See Figure 1 for more details.

To complete our discussion let us consider the inequality defined by a quadratic polynomial:



If and () the inequality (3) holds true for any value of (see Figure 1). If and () the inequality (3) is impossible for any value of ) (see Figure 1). If and () the inequality (3) holds true for all values of satisfying or , where are the roots of the quadratic equation (1) (see Figure 1). If and () the inequality (3) holds true for all values of satisfying , where are the roots of the quadratic equation (1) (see Figure 1). If If and then the inequality (3) holds true for all values of If and (except , see Figure 1). If If and then the inequality (3) is impossible for any value of and (see Figure 1).


Solution

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