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aama100

  • 4 years ago

(x-8)^2=36 find exact and approximate solutions to three decimal places. a) solve 8x+x(x-6)=0 b) find the x-intercepts of f(x)=8x+x(x-6) a)what are the solution x=? c) what are the x-intercepts

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  1. imranmeah91
    • 4 years ago
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    8x+x(x-6) or (x-8)^2=36 ?

  2. aama100
    • 4 years ago
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    a) (x-8)^2=36 find exact and approximate solutions to three decimal places. b) solve 8x+x(x-6)=0 c) find the x-intercepts of f(x)=8x+x(x-6) 1)what are the solution x=?

  3. myininaya
    • 4 years ago
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    a) take square root of both sides: x-8=pm 6 pm means plus or minus then add 8 on both sides

  4. saljudieh07
    • 4 years ago
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    x^2-16X+64=36 X^2-16X+28=0 Then, just find it using the quadratic formula, so the answers will be: (16+/-12)/2= 14 or 2 Myiniaya i think that way is incorrect because you are cancelling one of the roots, and you need all the answers

  5. polpak
    • 4 years ago
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    She said \(\pm 6\) which is not cancelling one of the roots.

  6. polpak
    • 4 years ago
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    \[(x-8)^2 = 36\]\[\implies x-8 = \pm6\]\[\implies x = 8\pm 6\]\[\implies x \in \{2,14\}\]

  7. saljudieh07
    • 4 years ago
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    for b) you have (8x+x)(x-6)=0 expand: 8x^2-48x+x^2-6x=0 simplify: 9x^2-54x=0 x=0 OR x= 5 and polpak, sorry didnt pay attention to the pm, so pm is +/-, and yes that would be correct!

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