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1. Find the product of all constants t such that the quadratic x^2 + tx - 10 can be factored in the form $(x+a)(x+b)$, where $a$ and $b$ are integers.

 

2.Find all real numbers t such that {2}/{3} t - 1 < t + 7 \le -2t + 15. Give your answer as an interval.

 May 11, 2019
 #1
avatar+7712 
+1

\(x^2+tx-10 = (x+a)(x+b)\\ x^2 + tx - 10 = x^2 + (a+b)x+ab\\ \therefore ab = -10\\ \text{For }a,b\in\mathbb Z,(a,b)=(10,-1),(5,-2),(2,-5),(1,-10),(-1,10),(-2,5),(-5,2),(-10,1)\\ \therefore t = 9,3,-3,-9\\ \text{Product of all t} = 9^2\cdot 3^2 = 729\)

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 May 11, 2019
 #2
avatar+103789 
+1

I'm assuming that this is

 

(2/3)t - 1 < t + 7 ≤ 2t + 15

 

We have two inequalities

(2/3)t - 1 < t + 7                         and            t + 7 ≤ 2t + 15

 

-8 < (1/3)t                                                    -8 ≤ t 

 

-24 < t

 

Taking the most restrictive interval  we have that

 

[-8, inf )

 

 

cool cool cool

 May 11, 2019

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