For what values of k does (k+1)x2−6x+5=0 have two equal real roots?

k =

Step-by-step explanation:

Using the determinant Δ = b² - 4ac to determine the value of k

The condition for the equation to have 2 equal real roots is

b² - 4ac = 0

Given

(k + 1)x² - 6x + 5 = 0 ← in standard form

with a = k + 1, b = - 6 and c = 5, then

(- 6)² - 4(k + 1) × 5 = 0, that is

36 - 20(k + 1) = 0 ← distribute and simplify left side

36 - 20k - 20 = 0

16 - 20k = 0 ( subtract 16 from both sides )

- 20k = - 16 ( divide both sides by - 20 )

k = =

well there are two answers for that.

1: you need 120 ones then that will equal up to 120.

2: you could also do one 100 and 2 tens.

those are the answers that i have thought of.