(x^2-2)^2-10(x^2-2)+21=0 find one value of x that is a solution to the equation
1 answer:
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Easy peasy
the bit where it says
D={something}
those are the numbers you should input for x to get y values
3.
first solve for y to make life easier
-3x-5y=20
times -1
3x+5y=-20
minus 3x both sides
5y=-3x-20
divide both sides by 5
y=-3/5x-4
sub value of the domain
x=-10
y=-3/5(-10)-4
y=6-4
y=2
a point is (-10,2)
x=-5
y=-3/5(-5)-4
y=3-4
y=-1
(-5,-1) is another
x=0
y=-3/5(0)-4
y=-4
(0,-4) is another
x=5
y=-3/5(5)-4
y=-3-4
y=-7
(5,-7)
points are (-10,2), (-5,-1), (0,-4), (5,-7)
4.
input
x=-2
y=(-2)^2-3
y=4-3
y=1
(-2,1)
x=-1
y=(-1)^2-3
y=1-3
y=-2
(-1,-2)
x=0
y=(0)^2-3
y=-3
(0,-3)
x=1
y=(1)^2-3
y=1-3
y=-3
(1,-3)
x=2
y=(2)^2-3
y=4-3
y=1
(2,1)
the points are (-2,1), (-1,-2), (0,-3), (1,-2), (-2,1)
see graph below
the bit where it says
D={something}
those are the numbers you should input for x to get y values
3.
first solve for y to make life easier
-3x-5y=20
times -1
3x+5y=-20
minus 3x both sides
5y=-3x-20
divide both sides by 5
y=-3/5x-4
sub value of the domain
x=-10
y=-3/5(-10)-4
y=6-4
y=2
a point is (-10,2)
x=-5
y=-3/5(-5)-4
y=3-4
y=-1
(-5,-1) is another
x=0
y=-3/5(0)-4
y=-4
(0,-4) is another
x=5
y=-3/5(5)-4
y=-3-4
y=-7
(5,-7)
points are (-10,2), (-5,-1), (0,-4), (5,-7)
4.
input
x=-2
y=(-2)^2-3
y=4-3
y=1
(-2,1)
x=-1
y=(-1)^2-3
y=1-3
y=-2
(-1,-2)
x=0
y=(0)^2-3
y=-3
(0,-3)
x=1
y=(1)^2-3
y=1-3
y=-3
(1,-3)
x=2
y=(2)^2-3
y=4-3
y=1
(2,1)
the points are (-2,1), (-1,-2), (0,-3), (1,-2), (-2,1)
see graph below