Answer in order:
False
True
False
False
False
False
False
True
True
Step-by-step explanation:
All of the solutions to the equation[tex]3x^2 - 12 = 0[/tex]are x = 12 and x = -12
Solve:
Common factor - [tex]3(x^2-4)=0[/tex]
Divide both sides
[tex]x^2-4=0[/tex]
Use the quadratic formula
[tex]x=\sqrt{4,x}=\sqrt{-4[/tex]
x = 2, x = -2
Thus, This is false..
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There are two unique solutions to the equations (x-3)^2 = 16
one variable in matrix = one possible value.
Therefore, x = 7, x = -1
Hence this is True
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The solutions for the equation 2(x-3)^2-18 = 0 are x = 6 and x = 0
[tex]2x^3-18x^2+54x-72=0[/tex]
[tex]x=5.080084[/tex]
Hence this is false
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The solutions for the equation 2(x-5)^2-8=0 are x = 7 and x = -7
Add both sides by 8
2(x-5)^2-8+8=0+8
2(x-5)^2=8
(x-5)^2=4
x = 7, x = 3
Hence, this is false
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The solutions for the equation (x + 3)^2-25 = -8 are x = 2 and x = -8
[tex](x+3)^2=17[/tex]
[tex]x=\sqrt{17-3,}=-\sqrt{17}-3[/tex]
Hence this is false.
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The solutions for the equation 2(2x-1)^2=18 are x = 5 and x = -4
[tex](2x-1)^2=9[/tex]
x=2, x = -1
Hence this is false.
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The only solution for equation (2x-1)^2-49=0 is x = 4
[tex](2x-1)^2=49[/tex]
x = 4, x = -3
Hence, this is false
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The solutions for the equation 3(x+2)^2 - 3 = 0 are x = -3 and x = -1
[tex]3(x+2)^2=3[/tex]
[tex](x+2)^2=1[/tex]
x = -1, x = -3
Hence, this is true
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The solutions for the equation 5x^2 - 180 = 0 are x = 6 and x = -6
Add 180 to both sides
[tex]5x^2-180+180=0+180[/tex]
[tex]5x^2 = 180[/tex]
[tex]x^2 = 36[/tex]
[tex]x=\sqrt{36},x=-\sqrt{36}[/tex]
x = 6, x = -6
Hence, this is true.
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[RevyBreeze]
