Question

(5b-4)(b-3) find zeros of this quadratic

Answer 1

To find the zeros of the quadratic (5b-4)(b-3), use the quadratic formula by substituting the given values, and then simplify to find the solutions.

Step-by-step explanation:

This expression is a quadratic equation of the form at² + bt + c = 0, where the constants are a = 5, b = -17, and c = 12. To find the zeros of this quadratic, we can use the quadratic formula:

x = (-b ± √(b² - 4ac)) / (2a)

Substituting the given values into the formula, we have:

x = (-(-17) ± √((-17)² - 4(5)(12))) / (2(5))

Simplifying further:

x = (17 ± √(289 - 240)) / 10

x = (17 ± √49) / 10

This gives us two solutions:

x = (17 + 7) / 10 = 24/10 = 2.4

x = (17 - 7) / 10 = 10/10 = 1

Answer 2

Answer:
`(4)/(5)`, 3

Step-by-step explanation:

1. Simplify: 5b2−19b+12=0

2. Factor by grouping: (5b−4)(b−3)=0

3. If any individual factor on the left side of the equation is equal to 0, the entire expression will be equal to 0: 5b−4=0, 5b-4=0b−3=0

4. Set 5b−45b-4 equal to 0 and solve for b: b=
`(4)/(5)`

5. Set b−3b-3 equal to 0 and solve for b: b=3
6. The final solution is all the values that make (5b-4)(b-3)=0 true: b =
`(4)/(5)`, 3