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kenny6666 [7]
3 years ago
8

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

Mathematics
1 answer:
Anna35 [415]3 years ago
5 0

Answer:

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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What is the expression in factored form?x^2+13x+14
nydimaria [60]

Answer:

\left(x-\dfrac{-13+\sqrt{113}}{2}\right)\left(x-\dfrac{-13-\sqrt{113}}{2}\right).

Step-by-step explanation:

To factor the expression x^2+13x+14 you need to know its roots.

First find the discriminant of this quadratic polynomial:

D=13^2-4\cdot 1\cdot 14=169-56=113.

Then the roots of the expression x^2+13x+14 are

x_{1,2}=\dfrac{-13\pm \sqrt{113}}{2}.

Now the factored form is

\left(x-\dfrac{-13+\sqrt{113}}{2}\right)\left(x-\dfrac{-13-\sqrt{113}}{2}\right).

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{Thanks for answering my questions!} #thanks
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