The quadratic equation $x^2-5x+t =3x$ has only positive integer roots. Find the average of all distinct possible values of $t$.

wiseowl Dec 10, 2023

#1**0 **

To solve for the roots of the quadratic equation, we can use the quadratic formula:

x = (-b +/- sqrt(b^2 - 4ac)) / 2a

In this case, a = 1, b = -5, and c = t - 3. Substituting these values into the formula, we get:

x = (5 +/- sqrt(25 - 4(1)(t - 3))) / 2

Simplifying the equation, we get:

x = (5 +/- sqrt(29 - 4t)) / 2

Since the quadratic equation has only positive integer roots, both x must be integers. Therefore, the discriminant (29 - 4t) must be a perfect square. Since t is an integer, the discriminant can only be a perfect square if 29 - 4t is a positive multiple of 4. This means that 29 - 4t has the form 4k, where k is an integer. Solving for t, we get:

t = (29 - 4k) / 4

Since t is an integer, k can only take on values from 1 to 7. This gives us 7 possible values of t:

t = 6, 5, 4, 3, 2, 1, 0

The average of these distinct values is:

(6 + 5 + 4 + 3 + 2 + 1 + 0) / 7 = 21 / 7 = 3

Therefore, the average of all distinct possible values of t is 3.

BuiIderBoi Dec 10, 2023

#2**+1 **

*The quadratic equation x ^{2} – 5x + t = 3x has only positive integer roots. Find the average of all distinct possible values of t.*

Subtract 3x from both sides x^{2} – 8x + t = 0

Note that the resulting

equation will factor. (x – 1)(x – 7) = 0 gives us t = 7

(x – 2)(x – 6) = 0 gives us t = 12

(x – 3)(x – 5) = 0 gives us t = 15

(x – 4)(x – 4) = 0 gives us t = 16

7 + 12 + 15 + 16 50

Average of possible t's = ––––––––––––––– = –––– = **12.5**

4 4

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Bosco Dec 11, 2023