What should be the value of b in x2+bx−55=0 so that its solutions are −11 and 5? (1 point)

Responses

−16
negative 16

6
6

−6
negative 6

16
16

1 answer

To find the value of \( b \) in the quadratic equation \( x^2 + bx - 55 = 0 \) given that its solutions are \( -11 \) and \( 5 \), we can use Vieta's formulas. According to Vieta's formulas, for a quadratic equation \( ax^2 + bx + c = 0 \), the sum of the solutions \( r_1 + r_2 \) is given by \( -\frac{b}{a} \) and the product of the solutions \( r_1 \cdot r_2 \) is given by \( \frac{c}{a} \).

  1. Sum of the roots: \[ r_1 + r_2 = -11 + 5 = -6 \] According to Vieta's: \[ -\frac{b}{1} = -6 \implies b = 6 \]

  2. Product of the roots: \[ r_1 \cdot r_2 = -11 \cdot 5 = -55 \] According to Vieta's: \[ \frac{-55}{1} = -55 \] This condition is satisfied with the given roots.

Thus, the value of \( b \) should be \( 6 \).

Therefore, the answer is: 6