find the coordinates of the points of intersection for the graphs of x^2+y^2=4 and y=2x-1
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OpenStudy (anonymous):
the first equation is that of a circle and the other one is that of a line...
so at most there are 2 intersections (or solutions to this system of equations....
substitute the second equation into the first equation...
OpenStudy (anonymous):
\[\large x^2+y^2=4,\space y=2x-1\rightarrow x^2+(2x-1)^2=4 \]
can you solve that?
OpenStudy (anonymous):
hmmm i dont know..sorry im really confused
OpenStudy (anonymous):
Can you expand this:
(2x−1)² = ....
OpenStudy (anonymous):
solve the equation:
\[\large x^2+(2x-1)^2=4 \]
\[\large x^2+(2x-1)(2x-1)=4\]
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OpenStudy (anonymous):
ok
OpenStudy (anonymous):
@kenneyfamily what do you mean by "ok" ???
OpenStudy (anonymous):
trying to follow
OpenStudy (anonymous):
(2x−1)² = ....
OpenStudy (anonymous):
hold on
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OpenStudy (anonymous):
4x^2-4x+1
OpenStudy (anonymous):
Great :)
x² + 4x² - 4x + 1 -4 = ....
OpenStudy (anonymous):
5x^2-4x-3
OpenStudy (anonymous):
Can you find intersection x by solving quadratic equation:
5x²-4x-3 = 0
OpenStudy (anonymous):
hint please? sorry
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OpenStudy (anonymous):
Δ = b² - 4ac
OpenStudy (anonymous):
b = -4, a = 5, c= -3
OpenStudy (anonymous):
ok and how do you find the intersection from that?
OpenStudy (anonymous):
x = ( -b ± √Δ ) / 2a
OpenStudy (anonymous):
Then plug x value into y = 2x -1 to get y value!
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OpenStudy (anonymous):
thank you!!!! so helpful
OpenStudy (anonymous):
@kenneyfamily I hope you look back your previous post to find my answers to learn!
It'll benefit for your further learning!