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The quadratic equation $x^2-mx+24 = 10$ has roots $x_1$ and $x_2$. If $x_1$ and $x_2$ are integers, how many different values of $m$ are possible?

 Apr 8, 2024

Best Answer 

 #2
avatar+128794 
+1

Simplify

 

x^2 - mx + 14  = 0

 

Factors of 14                                                m

(1 , 14)                   (x + 1) (x + 14)              -14

(-1, -14)                  (x - 1)  (x - 14)               14

(2,7)                       (x + 2) ( x + 7)                -9

(-2, -7)                   (x - 2) ( x - 7)                   9

 

cool cool cool

 Apr 9, 2024
 #1
avatar+214 
+1

The quadratic equation takes the form of x^2 - mx + 24 = 0. Given that x_1 and x_2 are its roots and are integers, it means that the sum of these roots (x_1 + x_2) equals m (based on Vieta's formulas), and the product of these roots equals 24 (x_1*x_2=24). In this case, we need to factor the constant term (24), in other words, list all possible pairs of integers that multiply to give 24. These include {1,24}, {-1,-24}, {2,12}, {-2,-12}, {3,8}, {-3,-8}, {4,6}, {-4,-6}, and their negative counterparts. As both roots are integers, and each pair of factors correspond to one unique value of m (the sum of those two integers), there are altogether 16 different values for m, considering both positive and negative pairs.

Answer is: 16

 

smileycoolsmiley

 #2
avatar+128794 
+1
Best Answer

Simplify

 

x^2 - mx + 14  = 0

 

Factors of 14                                                m

(1 , 14)                   (x + 1) (x + 14)              -14

(-1, -14)                  (x - 1)  (x - 14)               14

(2,7)                       (x + 2) ( x + 7)                -9

(-2, -7)                   (x - 2) ( x - 7)                   9

 

cool cool cool

CPhill Apr 9, 2024

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