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Answer No 14

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  21x² +20x +100 = 0

Step-by-step explanation:

We know the sum of the roots of x² +bx +c = 0 is -b, and their product is c. If the roots are α and β, then ...

The sum of the roots of the new equation will be ...

  -b' = (α+1/β)+(β+1/α) = (α+β) +(1/α +1/β) = (α+β)(1 +1/(αβ))

The product of the roots of the new equation will be ...

  c' = (α+1/β)(β+1/α) = αβ +2 +1/(αβ)

Using the above relations for (α+β) and αβ, we find that ...

  -b' = (-b)(1 +1/c)

  c' = c + 2 + 1/c

For the given equation, our definition of b and c is ...

  b = 2/3

  c = 7/3

so the new equation has values ...

  b' = (2/3)(1 + 1/(7/3) = (2/3)(10/7) = 20/21

  c' = 7/3 + 2 + 1/(7/3) = 13/3 + 3/7 = 100/21

So, the equation with the roots of interest is ...

  x² +20/21x +100/21 = 0

Multiplying by 21 gives ...

  21x² +20x +100 = 0

step-by-step explanation:

to determine whether a given function is even, we simply substitute x with -x in the expression. if the result after simplification is identical to the original equation, then the function is said to be even other wise odd.

f(-x) = (-x)^2 - (-x) + 8

      = x^2 + x +8

the resulting function is not identical to the original one;

f(-x) ≠ f(x)

therefore, f(x) is not an even function