MTEL 47 Pracice Test, Q 40: find x-coord. of soln to system y = 2x - 4, y = 2x^2 - 4x

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Saikyn

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Hello, I'm studying for the MTEL #47 Practice Test.

I'm having trouble solving problem number 40:
Problem 40: Use the systems of equations to answer the question that follows
y = 2x - 4
y = 2x^2 - 4x - 7

Which of the following expressions represents the x-coordinate of a point in the solution set of the above system of equations.

A. (1+2*sqrt(23)) / 2
B. (2 -sqrt (76)) / 4
C. (3 + (sqrt15)) / 2

The correct answer is in fact C, however, I'm not quite sure how it is obtained.

My strategy was to set the two original equations equal one another like so

2x - 4 = 2x^2 - 4x - 7

Subtract 2x from both sides

-4 = 2x^2 -6x - 7

Add 4 to both sides

0 = 2x^2 - 6x -3

now since A = 2 B = 6 C = -3 I thought you could just plug into the quadratic equation. I also tried to complete the square but that got really messy and I wans't sure how to deal with the square roots.

Any help or guideance with this would be appreciated - I've been struggling to figure it out.

Thanks in advance.
 
y = 2x - 4
y = 2x^2 - 4x - 7

Which of the following represents an x-coodinate of a point in the solution set of the above system of equations?

The solution is 3+(sqrt(15)/2 ccording to the document MTEL 47 #40 but how do they reach that? I ge really confused
 
Saikyn said:
Hello, I'm studying for the MTEL #47 Practice Test.

I'm having trouble solving problem number 40:
Problem 40: Use the systems of equations to answer the question that follows
y = 2x - 4
y = 2x^2 - 4x - 7

Which of the following expressions represents the x-coordinate of a point in the solution set of the above system of equations.

A. (1+2*sqrt(23)) / 2
B. (2 -sqrt (76)) / 4
C. (3 + (sqrt15)) / 2

The correct answer is in fact C, however, I'm not quite sure how it is obtained.

My strategy was to set the two original equations equal one another like so

2x - 4 = 2x^2 - 4x - 7

Subtract 2x from both sides

-4 = 2x^2 -6x - 7

Add 4 to both sides

0 = 2x^2 - 6x -3

now since A = 2 B = 6 C = -3 I thought you could just plug into the quadratic equation. I also tried to complete the square but that got really messy and I wans't sure how to deal with the square roots.
Click to expand...
I would have proceeded the exact same way. Using the Quadratic Formula is a way of dealing with those square roots.

One of the listed answers is indeed correct. Please reply showing what you got with the Quadratic Formula. Thank you! ;)
 
Saikyn said:
C. (3 + (sqrt15)) / 2

The correct answer is in fact C

...

0 = 2x^2 - 6x -3 . This is correct
Click to expand...

In case you got 3/2 + sqrt(15)/2, it's the same answer (different form).
 
stapel said:
I would have proceeded the exact same way. Using the Quadratic Formula is a way of dealing with those square roots.

One of the listed answers is indeed correct. Please reply showing what you got with the Quadratic Formula. Thank you! ;)
Click to expand...
So, if A = 2, B = -6, and C = -3

We get

6 plus or - sqrt(-6^2 - 4*2*-3) / (2*2)

= 6+ or - sqrt(36+24)/4

=6+ or - sqrt(60) / 4
?? I don't understna dhow you go from the inputs to the output :\ i'm not getting 3/2 which I know C is the answer.

I am just trying to understand how they got it of the form provided in the answers.
 
Saikyn said:
So, if A = 2, B = -6, and C = -3

We get

[6 plus or - sqrt{(-6)^2 - 4*2*(-3)}] / (2*2)

= 6+ or - sqrt(36+24)/4

=6+ or - sqrt(60) / 4
?? I don't understna dhow you go from the inputs to the output :\ i'm not getting 3/2 which I know C is the answer.

I am just trying to understand how they got it of the form provided in the answers.
Click to expand...

[6 plus or - sqrt(-6^2 - 4*2*-3)] / (2*2)



= [6 ±√(36 - 4*2*(-3)] / 4

= [6 ±√60] / 4

= [6 ±2√15] / 4

= [3 ± √15] /2
 
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