What should be the value of b in x2+bx−55=0

so that its solutions are −11
and 5? (1 point)
Responses

6
6

−16
negative 16

−6
negative 6

16

1 answer

To find the value of \( b \) in the quadratic equation \( x^2 + bx - 55 = 0 \) given that the solutions are \( -11 \) and \( 5 \), we can use Vieta's formulas.

According to Vieta's formulas, for a quadratic equation of the form \( x^2 + bx + c = 0 \):

  • The sum of the roots \( r_1 + r_2 = -b \)
  • The product of the roots \( r_1 \cdot r_2 = c \)

In our case, the roots are \( r_1 = -11 \) and \( r_2 = 5 \).

  1. Calculate the sum of the roots: \[ -11 + 5 = -6 \] According to Vieta's, this gives us: \[ -b = -6 \quad \Rightarrow \quad b = 6 \]

  2. Calculate the product of the roots: \[ -11 \cdot 5 = -55 \] This verifies the constant term \( c = -55 \).

Thus, the value of \( b \) that results in the roots of \( -11 \) and \( 5 \) for the equation \( x^2 + bx - 55 = 0 \) is:

\[ \boxed{6} \]