Question

9.02] What are the solutions of x2 + 8x − 20 = 0?

x = 2, x = 10

x = −2, x = −10

x = 2, x = −10

x = −2, x = 10

Answer 1

The answer would be the second to last one because -2 * 10 = -20 ans -2+10 = 8
put that into brackets as (x-2) (x+10).....
so x=2 as 2-2 =0, x=-10 as -10+10 = 0

Answer 2

Oh, okay, thx! thx for the explanation!

Answer 3

no problem

Answer 4

:D

Answer 5

Could you help me with one more?

Answer 6

[9.01] What is the vertex for the graph of y = −2x2 + 8x − 13?

(2, 11)

(2, −5)

(−2, −37)

(−2, −21)

Answer 7

I'm not too sure about this one it requires a graph and i cant picture it

Answer 8

Oh, okay, then...this one?

Answer 9

9.01] When the quadratic equation y = (x − 4)(2x + 5) is written in standard form, what is the value of the coefficient "b"?

2

3

−3

−20
is it a?

Answer 10

this one you have to expand the brackets so it becomes, 2x^2 - 8x - 20 + 5x
simplify it to be ... 2x^2 - 3x -20
so the answer would be -3

Answer 11

For the second question: y = −2x2 + 8x − 13 = -(x-2)^2- 5
Therefore, vertex will be (2, -5).

Answer 12

Sorry typo ....
For the second question: y = −2x2 + 8x − 13 = -2(x-2)^2- 5
Therefore, vertex will be (2, -5).

Answer 13

could you show how you got that without the graph, i would like to know

Answer 14

We have an equation ax2 + bx + c =0
The vertex is \[V=(\frac{ -b }{ 2a },c-\frac{ b^{2} }{ 4a })\]

Answer 15

Hint: perfect squere
y = −2x2 + 8x − 13 = -2(x-2)^2- 5 = 0

x-2 = 0, x =2

y = -5

V = (2, -5)

Answer 16

OMG, thank you so much! LOL, it was so nice of you to help!