Question

If there is only one solution to the equation 0=-x^2 + bx -8 what are the possible values of b?

Answer 1

Answer:

$b = 4\sqrt{2}$, $b = -4\sqrt{2}$

Step-by-step explanation:

For a single solution, you must have a perfect square trinomial.

$-x^2 + bx - 8 = 0$

$x^2 - bx + 8 = 0$

$(x - \sqrt{8})^2 = 0$

$x^2 - 2\sqrt{8}x + 8 = 0$

$x^2 - 4\sqrt{2}x + 8 = 0$

$-x^2 + 4\sqrt{2}x - 8 = 0$

$b = \pm 4\sqrt{2}$