Quadratic and linear Intersection

AI Thread Summary
The discussion revolves around finding the points of intersection between the line y = x and the parabola y = 6x - x² - 10. The user initially attempted to solve the equation but received an incorrect mark from the teacher. They used the quadratic formula's discriminant (b² - 4ac) to determine the number of intersections but faced confusion regarding their calculations. After some back and forth, the user clarified that they had mistakenly typed their answer and confirmed it was A (none). The thread highlights the importance of clear communication and accurate calculations in solving quadratic equations.
master_333
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Homework Statement


How many points of intersection does the line y = x have with the parabola y = 6x − x2 −10 ?
A) none
B) one
C) two
D) three
E) more than three

Homework Equations

The Attempt at a Solution


x = 6x - x^2 - 10
-x^2+5x - 10
Used b^2 - 4ac to check the number of intersections.

The answer I got is . However my teacher marked it as incorrect. I still cannot find the problem, please help. If my teacher marked it wrong, please tell me so because I think he did. Thanks.
 
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master_333 said:

The Attempt at a Solution


x = 6x - x^2 - 10
-x^2+5x - 10
Used b^2 - 4ac to check the number of intersections.

The answer I got is . However my teacher marked it as incorrect. I still cannot find the problem, please help. If my teacher marked it wrong, please tell me so because I think he did. Thanks.

What was your answer? What did you plug in for b2 - 4ac? Your approach is correct, but I'd draw a graph to be sure. (Hint: you know the parabola is negative, and you know the y-intercept is -10.)
 
master_333 said:

Homework Statement


How many points of intersection does the line y = x have with the parabola y = 6x − x2 −10 ?
A) none
B) one
C) two
D) three
E) more than three

Homework Equations

The Attempt at a Solution


x = 6x - x^2 - 10
-x^2+5x - 10
Where did the = go? You have it two lines above, but not in the line above here.
master_333 said:
Used b^2 - 4ac to check the number of intersections.

The answer I got is .
Is what?
master_333 said:
However my teacher marked it as incorrect.
Your teacher marked what as incorrect?
You need to give us enough information so that we can provide meaningful help.
master_333 said:
I still cannot find the problem, please help. If my teacher marked it wrong, please tell me so because I think he did. Thanks.
 
Sorry I know the right answer now. However I typed it in wrong. I forgot to write that the answer i got is A . Thanks
 
master_333 said:
Sorry I know the right answer now. However I typed it in wrong. I forgot to write that the answer i got is A . Thanks
Are you saying that your teacher's answer is not A, or that you typed the question wrongly (or both)?
 

Thread 'Find the polar form of a complex number'

The working out suggests first equating ## \sqrt{i} = x + iy ## and suggests that squaring and equating real and imaginary parts of both sides results in ## \sqrt{i} = \pm (1+i)/ \sqrt{2} ## Squaring both sides results in: $$ i = (x + iy)^2 $$ $$ i = x^2 + 2ixy -y^2 $$ equating real parts gives $$ x^2 - y^2 = 0 $$ $$ (x+y)(x-y) = 0 $$ $$ x = \pm y $$ equating imaginary parts gives: $$ i = 2ixy $$ $$ 2xy = 1 $$ I'm not really sure how to proceed from here.
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