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Find the greatest value of $a$ such that $a^2-10a+21 \le 0$.

 Oct 26, 2014

Best Answer 

 #2
avatar+130466 
+5

Setting the polynomial = 0 and factoring, we have

(a -7)(a-3) = 0  ..... so a = 7 or a = 3

And for all a > 7, the polynomial is postive.

So, the greatest "a" that makes the polynomial ≤ 0 is when a = 7.

 

 Oct 26, 2014
 #1
avatar+23254 
+5

If you change the problem to this:  a² - 10a + 21  =  0

Then, factoring:                            (a - 7)(a - 3)  =  0    --->  a = 3 or a = 7.

Going back to your original problem, an inequalty:  ≤ 0:

If you try any number (in the original problem) smaller than 3, your answer is positive; they don't work.

If you try any number from 3 through 7, your answer is negative; they all work, with the largest number = 7.

If you try any number larger than 7, your answer is positive; they don't work.

 Oct 26, 2014
 #2
avatar+130466 
+5
Best Answer

Setting the polynomial = 0 and factoring, we have

(a -7)(a-3) = 0  ..... so a = 7 or a = 3

And for all a > 7, the polynomial is postive.

So, the greatest "a" that makes the polynomial ≤ 0 is when a = 7.

 

CPhill Oct 26, 2014

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