Question

Let f(x) = x2 + 6x + c for all real number x, where c is a real number A value of C for which f(1/3) = 0
have exactly 3 distinct real solution is
(where a, b e N), then find the value of la-bl​

Answer 1

living things get rid

is the prase and is the sentence

Answer 2

Answer:

The value of la-bl​ = 8.6

Step-by-step explanation:

• Finding the value of c: The given polynomial is: f(x) = x² + 6x + c , where c is a real number.

According to the given condition when x = $\frac{1}{3}$ , then f($\frac{1}{3}$) = 0

Putting the value of x in above equation we get,  $\frac{1}{3}$² + 6($\frac{1}{3}$) + c = 0

                                                                                $\frac{1}{9}$ + 2 + c = 0

                                                                                 c + $\frac{19}{2}$ = 0

                                                                                 c = - $\frac{19}{2}$ = -9.5

• Solve the equation: From the equation, x² + 6x - 9.5 = 0

                                                                    x = $\frac{-6 +\sqrt{6^2+4*9.5} }{2}$ or  $\frac{-6 -\sqrt{6^2+4*9.5} }{2}$

                                                                   x = 1.3 or -7.3

  The two values are 1.3 and - 7.3.

  So, a = 1.3 and b = -7.3

  a - b = 1.3 - (- 7.3) = 1.3 + 7.3 = 8.6

So, la-bl​ = 8.6

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