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If the polynomial x^2+bx+c has exactly one real root and b=c+11, find the value of the product of all possible values of c.

 Dec 27, 2020
 #1
avatar+66 
+3

for the polynomial to have 1 real root, its discriminant, \(\sqrt{b^2-4ac}\), must be equal to zero.

In this case, it would mean that\(\sqrt{(c+11)^2-4c}\) must be equal to zero. We can then set up an equation:

\(\sqrt{(x+11)^2-4x}=0\\ (x+11)^2-4x=0\\ x^2+22x+121-4x=0\\ x^2+18x+121=0\)

(i replaced the variable c with x to make it less confusing)

the product of the roots of a quadratic is equal to the constant term divided by the coefficient of the quadratic term. In this case, it would be equal to 121/1 = 121. 

Therefore, the answer to your question is c = 121.

 Dec 27, 2020
 #2
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+1

....BUT   since b = c+ 11     b would be    132 under your scenario

x^2 + 132 x + 121    has 2 roots.....   hmmmmm.....

Guest Dec 27, 2020
 #3
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-2

From textor answer:

x2 + 18x + 121 = 0   gives the quadratic for solving for c

   using quadratic formula    c = - 9/4 +- 1/4 sqrt (403)i

                                                     the product of these two values

                                                         81/16 + 1/16  *(403) = 30.25

 Dec 27, 2020
 #4
avatar+112028 
+2

Answer 1 is correct but there is a typo.

 

It is not c=121

 

It is the product of all possible c is 121

 

\(c=-9\pm\sqrt{40}\;i\)

 Dec 28, 2020

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