How many distinct positive integer-valued solutions exist to the equation (x2 - 7x + 11)(x2 - 13x + 42) = 1

The given equation has TWO positive integer valued solutions, {6, 7}

Step-by-step explanation:

Here we are given two trinomial factors, each of which needs to be set equal to zero and in each case the resulting quadratic equation solved.

(x^2 - 7x + 11) has the coefficients {1, -7, 11}, and so the discriminant of this quadratic is b^2 - 4(a)(c), or 49 - 4(11), or 5.

Because this discriminant is positive, we know immediately that this quadratic has two real, unequal roots involving √5 (NO integer roots).

Next we focus on (x^2 - 13x + 42).  The discriminant is b^2 - 4(a)(c), or

169 - 168, or 1.  Again we see that there are two real, unequal roots:

      +13 ± √1                  +13 ± 1

x =  or  x =

             2                            2

OR x = 14/2 = 7 (integer) or x = 12/2 = 6 (integer).

The given equation has TWO positive integer valued solutions, {6, 7}

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