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    Which question are you troubled by, what have you tried to solve it?
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    (Original post by gdunne42)
    Which question are you troubled by, what have you tried to solve it?

    questions 1 , 2 , 9 and10
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    (Original post by stalleyzim)
    questions 1 , 2 , 9 and10
    Q1 and Q3 - since these cubics have a repeated root, you can denote this root by x=\alpha, and then note that f(\alpha) = 0. Then consider the derivative of these cubics. At x=\alpha you must have f'(\alpha) = 0. This gives another equation which you can use with the first and solve for the value of \alpha

    Q2 - Simply sub x=0 and show that the equality does not hold. Dividing it through by x^2 and rearranging it a bit yields 5(x^2+x^{-2})-16(x+x^{-1})-42=0. So given the substitution, express this eq. in terms of y. The only bit of work you really need to do here is work out what x^2+x^{-2} is in terms of y

    Q9 - begin by squaring both sides to yield 4x+2\sqrt{(3x+1)(x-1)}=7x+1 then rearrange for the root and square once more to get rid off it. Solve the quadratic at hand. Disregard solutions (if any) for which any of the expressions 3x+1, x-1, or 7x+1 are negative.

    Q10 - You can sketch both sides of the equation on the same graph and work from there. Otherwise, note that |x| \equiv \sqrt{x^2} so really you're just dealing with the eq. \sqrt{x^2}=3-\sqrt{(1-x)^2} which can be dealt in a similar fashion as Q9. OR yet another alternative - search for solutions in the regions x<0, 0<x<1 and x>1 individually by noting, for example, that for x<0 we have |x|=-x and |1-x|=1-x, hence the eq. for this region is just -x=3-(1-x)
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    thank you
 
 
 
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