To solve: the system of equations both algebraically and graphically.
The value of intersection points are
(y=±√√161+4732,−1+√16116)(±√47−√16132,−1−√16116) .
Given information: The given system of an equations is:
x2+4y2=4y=2x2−3
Calculation:
The value of x2
can be calculated from an equation (2):
x2+4y2=4 ........(1)y=2x2−3y+3=2x2y+32=x2 ........(2)
Substitute the value of equation (2) in equation (1):
y+32+4y2=4y+3+8y22=4y+3+8y2=8y+3+8y2−8=08y2+y−5=0
Using the
y=−b±√b2−4ac2ay=−1±√1−4×(−5)×82×8y=−1±√1+16016y=−1±√16116y=−1+√16116,y=−1−√16116
If the value of y=−1+√16116 then the value of,
x2=−1+√16116+32x2=−1+√161+4832x2=−1+√161+4832x2=√161+4732x=±√√161+4732
If the value of y=−1−√16116 , then the value of
x2=√−1−√16116+32x2=−1−√161+4832x2=47−√16132x2=−√161+4732x=±√47−√16132
Thus, the intersection points are
(±√√161+4732,−1+√16116)(±√47−√16132,−1−√16116)
Which are equivalent to (±1.3666, 0.731) and (±1.035, −0.856) .
Here, the graphical representation is as follow:
Graphical Interpretation:
In the graph the red curve represents x2+4y2=4 and blue curve represents y=2x2−3 .And from the graph it is observed that the curves are intersecting at 4 points, (±1.3666, 0.731) and (±1.035, −0.856) .
Chapter 8 Solutions
PRECALCULUS:GRAPHICAL,...-NASTA ED.
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