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AS Maths Help: Solving equations with roots/surds watch

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    Hi there, i could do with a little bit of help here as i want to know the method of how to do these 2 questions:
    Im sure this is quite simple in reality but i havent done these before so i would be grateful to know the method and the answer to these 2 questions please. Thanks.
    "Solve the following equations"
    1) t-5\surd t -14=0


    2) x-3\surd x =4
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    Make a substitution u = \sqrt{t} and see what happens.
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    make substitution then solve.
    first one t= 49
    second one x= 16
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    Sorry but im still a bit unsure what to do. If someone could show the full working out even just for the first one, i will probably be able to do the second on my own. Thanks.
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    (Original post by ~Sparked~)
    Sorry but im still a bit unsure what to do. If someone could show the full working out even just for the first one, i will probably be able to do the second on my own. Thanks.
    I'll give you a different example.

    Say you have r + \sqrt{r} - 2 = 0. First notice that r=(\sqrt{r})^2, so we can write this as
    (\sqrt{r})^2 + \sqrt{r} - 2 = 0

    Making the substitution u=\sqrt{r} then turns this equation into

    u^2+u-2=0

    This is a quadratic equation, so you can factorise it using whatever method you like; and this will give:

    (u-1)(u+2)=0

    ...which has roots u=1,\ u=-2. This isn't the final answer though, because we need our roots to be in terms of r; but we know that u=\sqrt{r}, and so putting this back in, we get

    \sqrt{r} = 1 or \sqrt{r} = -2

    The first of these would mean that r=1, but beware of the second: when we write \sqrt we implicitly use the positive root; and so there are no (real) numbers which have -2 as their principal square root, and so the only solution to the equation r+\sqrt{r}-2=0 is r=1.
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    Thanks, that clears things up.
 
 
 

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