Answer/Step-by-step explanation:
All of the solutions to the equation 3x^2 - 12 = 0 are x = 12 and x = -2
Answer: False
Explanation:
[tex]3x^2-12+12=0+12[/tex]
[tex]3x^2=12[/tex]
[tex]\frac{3x^2}{3}=\frac{12}{3}[/tex]
[tex]x^2=4[/tex]
[tex]\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}[/tex]
[tex]x=\sqrt{4},\:x=-\sqrt{4}[/tex]
[tex]x=2,\:x=-2[/tex]
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There are two unique solutions to the equations (x-3)^2 = 16
Note: Each variable in the matrix can have only one possible value, and this is how you know that this matrix has one unique solution.
Answer: True
Explanation:
[tex]\mathrm{For\:}\left(g\left(x\right)\right)^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}g\left(x\right)=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}[/tex]
[tex]x=7,\:x=-1[/tex]
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Ths solutions for the equation 2(x-3)^3 - 18 = 0 are x = 6 and x = 0
Answer: False
Explanation:
[tex]2\left(x-3\right)^3-18+18=0+18[/tex]
[tex]2\left(x-3\right)^3=18[/tex]
[tex]\frac{2\left(x-3\right)^3}{2}=\frac{18}{2}[/tex]
[tex]\left(x-3\right)^3=9[/tex]
[tex]\mathrm{For\:}g^3\left(x\right)=f\left(a\right)\mathrm{\:the\:solutions\:are\:}g\left(x\right)=\sqrt[3]{f\left(a\right)},\:\sqrt[3]{f\left(a\right)}\frac{-1-\sqrt{3}i}{2},\:\sqrt[3]{f\left(a\right)}\frac{-1+\sqrt{3}i}{2}[/tex]
[tex]=\sqrt[3]{9}+3,\:x=\frac{6-\sqrt[3]{9}}{2}+i\frac{\sqrt[3]{9}\sqrt{3}}{2},\:x=\frac{6-\sqrt[3]{9}}{2}-i\frac{\sqrt[3]{9}\sqrt{3}}{2}[/tex]
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~The solutions for the equation 2(x-5)^2-8=0 are x = 7 and x = -7
Answer: False
Explanation:
[tex]2\left(x-5\right)^2-8+8=0+8[/tex]
[tex]2\left(x-5\right)^2=8[/tex]
[tex]\frac{2\left(x-5\right)^2}{2}=\frac{8}{2}[/tex]
[tex]\left(x-5\right)^2=4[/tex]
[tex]\mathrm{For\:}\left(g\left(x\right)\right)^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}g\left(x\right)=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}[/tex]
[tex]x=7,\:x=3[/tex]
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The solutions for the equation (x + 3)^2-25 = -8 are x = 2 and x = -8
Answer: False
Explanation:
[tex]\left(x+3\right)^2-25+25=-8+25[/tex]
[tex]\left(x+3\right)^2=17[/tex]
[tex]\mathrm{For\:}\left(g\left(x\right)\right)^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}g\left(x\right)=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}[/tex]
[tex]x=\sqrt{17}-3,\:x=-\sqrt{17}-3[/tex]
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The solutions for the equation 2(2x-1)^2=18 are x = 5 and x = -4
Answer: False
Explanation:
[tex]\frac{2\left(2x-1\right)^2}{2}=\frac{18}{2}[/tex]
[tex]\left(2x-1\right)^2=9[/tex]
[tex]\mathrm{For\:}\left(g\left(x\right)\right)^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}g\left(x\right)=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}[/tex]
[tex]x=2,x=-1[/tex]
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The only solution for equation (2x-1)^2-49=0 is x = 4
Answer: False
Explanation:
[tex]\left(2x-1\right)^2-49+49=0+49[/tex]
[tex]\left(2x-1\right)^2=49[/tex]
[tex]\mathrm{For\:}\left(g\left(x\right)\right)^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}g\left(x\right)=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}[/tex]
[tex]x=4,\:x=-3[/tex]
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The solutions for the equation 3(x+2)^2 - 3 = 0 are x = -3 and x = -1
Answer: True
Explanation:
[tex]3\left(x+2\right)^2-3+3=0+3[/tex]
[tex]3\left(x+2\right)^2=3[/tex]
[tex]\frac{3\left(x+2\right)^2}{3}=\frac{3}{3}[/tex]
[tex]\left(x+2\right)^2=1[/tex]
[tex]\mathrm{For\:}\left(g\left(x\right)\right)^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}g\left(x\right)=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}[/tex]
[tex]x=-1,\:x=-3[/tex]
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The solutions for the equation 5x^2 - 180 = 0 are x = 6 and x = -6
Answer: True
Explanation:
[tex]5x^2-180+180=0+180[/tex]
[tex]5x^2=180[/tex]
[tex]\frac{5x^2}{5}=\frac{180}{5}[/tex]
[tex]x^2=36[/tex]
[tex]\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}[/tex]
[tex]x=\sqrt{36},\:x=-\sqrt{36}[/tex]
[tex]x=6,\:x=-6[/tex]
~Lenvy~