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    Name:  Screenshot_2016-04-01-13-36-04.png
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Size:  42.7 KB can anyone go through this question step by step? I don't get how to solve it. Thanks.
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     \displaystyle = \sqrt 2x^3 \times \frac{\sqrt{x^2}}{\sqrt{32}} .

     \sqrt{32}= ..

     \sqrt{x^2}=... <---- be careful with this one.
    Spoiler:
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     \sqrt{x^2} \neq x
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    (Original post by coconut64)
    Name:  Screenshot_2016-04-01-13-36-04.png
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Size:  42.7 KB can anyone go through this question step by step? I don't get how to solve it. Thanks.
    Step 1: a \div \frac{b}{c}  = a \times \frac{c}{b}: this gives \sqrt{2}(x^3) \div \sqrt{\frac{32}{x^2}} = \sqrt{2}(x^3) \times \sqrt{\frac{x^2}{32}}

    Step 2: \sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}, this gives: \sqrt{2}(x^3) \times \sqrt{\frac{x^2}{32}} = \sqrt{2}(x^3) \times \frac{\sqrt{x^2}}{\sqrt{32}}

    Step 3: \sqrt{32} = \sqrt{16 \times 2} = \sqrt{16}\sqrt{2} = 4\sqrt{2} so that \sqrt{2} \times \frac{1}{\sqrt{32}} = \frac{\sqrt{2}}{4\sqrt{2}} = \frac{1}{4}

    Step 4: \sqrt{2}(x^3) \times \frac{\sqrt{x^2}}{\sqrt{32}} = \frac{1}{4} \times x^3 \times \sqrt{x^2}

    Step 5: \sqrt{a} = a^{1/2} so \sqrt{x^2} = (x^{2})^{1/2}

    Step 6: (a^b)^c = a^{bc} so \sqrt{x^3} = (x^2)^{1/2} = x^{2 \times 1/2} = x^{1} = x

    Can you finish off from here?
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    (Original post by Zacken)
    Step 1: a \div \frac{b}{c}  = a \times \frac{c}{b}: this gives \sqrt{2}(x^3) \div \sqrt{\frac{32}{x^2}} = \sqrt{2}(x^3) \times \sqrt{\frac{x^2}{32}}

    Step 2: \sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}, this gives: \sqrt{2}(x^3) \times \sqrt{\frac{x^2}{32}} = \sqrt{2}(x^3) \times \frac{\sqrt{x^2}}{\sqrt{32}}

    Step 3: \sqrt{32} = \sqrt{16 \times 2} = \sqrt{16}\sqrt{2} = 4\sqrt{2} so that \sqrt{2} \times \frac{1}{\sqrt{32}} = \frac{\sqrt{2}}{4\sqrt{2}} = \frac{1}{4}

    Step 4: \sqrt{2}(x^3) \times \frac{\sqrt{x^2}}{\sqrt{32}} = \frac{1}{4} \times x^3 \times \sqrt{x^2}

    Step 5: \sqrt{a} = a^{1/2} so \sqrt{x^2} = (x^{2})^{1/2}

    Step 6: (a^b)^c = a^{bc} so \sqrt{x^3} = (x^2)^{1/2} = x^{2 \times 1/2} = x^{1} = x

    Can you finish off from here?
    Name:  1459515459784207341171.jpg
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Size:  235.6 KB I think this is right?
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    (Original post by coconut64)
    Name:  1459515459784207341171.jpg
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Size:  235.6 KB I think this is right?
    Perfect. First class work.
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    (Original post by coconut64)
    Name:  1459515459784207341171.jpg
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Size:  235.6 KB I think this is right?
    hmmm. try x=-1 and see if this works
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    (Original post by B_9710)
     \displaystyle = \sqrt 2x^3 \times \frac{\sqrt{x^2}}{\sqrt{32}} .

     \sqrt{32}= ..

     \sqrt{x^2}=... <---- be careful with this one.
    Spoiler:
    Show
     \sqrt{x^2} \neq x
    Thanks
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    (Original post by coconut64)
    Thanks
    I have to say, you did not get the right answer. As I said try any negative value of x and see if you get the same answer from the original expression and the final one you posted.
    It is right if there is a condition stated somewhere that  0 \leqslant x .
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    (Original post by coconut64)
    Thanks
    Would the question happen to have a condition x&gt;0 that you haven't shown us? I'd think so.
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    (Original post by Zacken)
    Would the question happen to have a condition x&gt;0 that you haven't shown us? I'd think so.
    Not really. Name:  1459516221392-1339212355.jpg
Views: 135
Size:  427.0 KB
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    (Original post by B_9710)
    I have to say, you did not get the right answer. As I said try any negative value of x and see if you get the same answer from the original expression and the final one you posted.
    It is right if there is a condition stated somewhere that  0 \leqslant x .
    True but it works with a positive number tho. This is a non calculator exam so I would probs to go along with it. The question doesn't give other info BTW. Thanks.
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    (Original post by coconut64)
    Not really. Name:  1459516221392-1339212355.jpg
Views: 135
Size:  427.0 KB
    Oh well, it's just crappy question writing on Edexcel's part, nothing you need worry about at this stage.

    Spoiler:
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    The reason why this is technically incorrect is that \sqrt{x^2} = |x| \neq x.
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    I'm saying the answer is  \displaystyle \frac{1}{4}x^3|x| .
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    (Original post by Zacken)
    Oh well, it's just crappy question writing on Edexcel's part, nothing you need worry about at this stage.
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    The reason why this is technically incorrect is that \sqrt{x^2} = |x| \neq x.
    Oh that I have abosolutely no clue what it is. I don't do further maths. Thanks again.
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    (Original post by coconut64)
    Oh that I have abosolutely no clue what it is. I don't do further maths. Thanks again.
    No problem.
 
 
 
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