# Sage said that the Quadratic Formula does not always work. Sage used it to solve the equation x2 – 3x - 2 = - 4, with a = 1,

###### Question:

c=-2. The formula gave x =

2

as the solutions to the equation. When Sage checked, neither one of them satisfied the equation.

How could you convince Sage that the Quadratic Formula does always work?

Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

O A Remind Sage to make sure the equation has a 0 on one side before trying to use the Quadratic Formula. Doing so gives a value of

instead of -2 for c. With the correct values of a, b, and c, the solutions will check.

ов.

Point out that Sage made an error in checking her solutions. Both

3+ /17 3- V17

2

2

do, in fact, satisfy the equation

x²-3x-2=-4.

and

O C. Find the error in Sage's calculation. Using the quadratic formula with a = 1, b = -3, and c= - 2 should have given solutions of

x=, and these will check.

(Type an exact answer, using radicals and i as needed. Use a comma to separate answers as needed.)

## Answers

1 is B.

2 is A

3 is B

graph the equation and check to see if the answers match any intercepting points.

3x^2+4x+2

please see the attached picture for full solution

hope it helps

Good luck on your assignment..

[tex]What is the quotient (3x^3+10x^2+10x+4) divided by (x+2)? Possible answers:3x^2+16x+423x^2+4x+23x^2[/tex]

4

Step-by-step explanation:

2+2=4

5x^2-x-4

Step-by-step explanation:

Combine the terms.

[tex]3x + 2 < - 4 \\ 3x < - 4 - 2 \\ 3x < - 6 \\ x < \frac{( -6)}{3} \\ \boxed{x < - 2}[/tex]

x<-2 is the right answer.B, A, B.

Step-by-step explanation:

Using a substitution method, it can be shown which (x,y) point does not satisfy the given equations.

1. Starting from the first function, represented algebraically by the equation 3x^2+6-4:

f(x)= -3x^2+6x-4

f(6)= -3(6)^2+6(6)-4

f(6)= (-3)(36)+36-4

f(6)= -76

That's why the point (x,y)= (6,-70) is not on the graph of y=-3x^2+6x-4

The point that is on the graph is (x,y)= (6,-76)

The answer is B. (6,-70)

2. For the second function, represented algebraically by the equation 2x^2+3x-10:

f(x)= 2x^2+3x−10

f(4)= 2(4)^2+3(4)-10

f(4)= 32+12-10

f(4)= 34

That's why the point (x,y)= (4,30) is not on the graph of y= 2x^2+3x−10

The point that is on the graph is (x,y)= (4,34)

The answer is A. (4,30)

3. For the last function, represented algebraically by the equation 1/2x^2-3x+7:

f(x)= −1/2 x^2 −3x+7

f(6)= −1/2 (6)^2 −3(6)+7

f(6)= −1/2 (36)-18+7

f(6)= -18-18+7

f(6)= -29

That's why the point (x,y)= (6,-30) is not on the graph of y= −1/2 x^2 −3x+7

The point that is on the graph is (x,y)= (6,-29)

The answer is B. (6,-30)

try to see it is pretty ez

[tex]What is the ananswer to 3x^2+2x-4=[/tex]

[tex]What is the ananswer to 3x^2+2x-4=[/tex]

Actually i was going to ask you the same question