Recent content by danielw

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    MHB Rationalising Surds: Get the Answers You Need - Daniel

    So since $$y^2=-(51)$$, $$y$$ is between 7 ($$7^2=49$$) and 8 ($$8^2=64$$), $$y=7$$?
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    MHB Rationalising Surds: Get the Answers You Need - Daniel

    Hi All This is my question. I don't know how to begin working on it. I already tried simplifying the first part to: $$ \sqrt{63} = \sqrt{7 \cdot 3^3}=\sqrt{7}\sqrt{3^2}=3\sqrt{7}$$ But this doesn't get me closer to answering the first part of the question, and I think the same technique...
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    MHB Simplification of surds and powers

    Hi All I have this problem. My workings are attached. My answer was: x=4 y=5 z=6 This is wrong. I don't know where I'm going wrong. I'd be really grateful if someone could help. Thanks! Daniel
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    MHB For which integers x,y is (4-6*sqrt(2))^2 = x+y*sqrt(2)?

    Hi All I have the following question. I have reviewed my notes but have not been able to crack this. I tried two different ways, both wrong. First: $$ (4-6*\sqrt2)^2=$$ $$16-24*\sqrt2-24*\sqrt2+(36*2) = 88-218*\sqrt2$$ so, $x=88$ and $y=218$My second method was $$(4-6*\sqrt2)^2=...
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    MHB Find unknowns in equation going from perfect square? to quadratic format?

    Hi all I'm trying to work out how to answer this type of problem. $$(6x+2z)^2-64=(ax+2z+8)(-8+bx+cz)$$ where a, b and c > 0 I have attempted the problem by expanding the brackets: $$=36x^2+24xz+4z^2-64$$ This is the same as $$(6x+2z)^2-(8)^2$$ Then subtracting from either 'side' of the...
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