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    Hi,

    I'm a bit stuck on the procedure on finding the answer to x^2 > 16 and x^2 > 36 and similar. Can someone help me out? I understand that square routing gives + or - but stuck after that.

    Thanks
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    Sketch the graph of y=x^2, then the rest should follow
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    Is there not a quicker way to solve it? It's just an inequality right?
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    (Original post by ShivP_98)
    Is there not a quicker way to solve it? It's just an inequality right?
    With practice, sketching the graphs to solve inequalities of the form ax^2 + bx + c > 0 is trivial and takes only a few seconds to roughly scrawl the sketch out on a piece of paper. The main thing is identifying the roots of the quadratic and sketching the positive and/or negative shape of the quadratic with respect to the roots. The solution to the inequality follows immediately after.

    If you want, you could visualise the graph in your head and solve it, but I would recommend against that.

    In your case, simply sketch out the graph of (x-4)(x+4), with roots at \pm 4 and see for what values of x for which the graph of (x-4)(x+4) is above the x-axis. Is it for x-values between -4 and 4, bigger than 4, less than 4, both bigger and less?
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    (Original post by ShivP_98)
    Hi,

    I'm a bit stuck on the procedure on finding the answer to x^2 > 16 and x^2 > 36 and similar. Can someone help me out? I understand that square routing gives + or - but stuck after that.

    Thanks
    The zeros occur at 4 and -4
    So you consider x values either side of those
    You will need to be either between or outside of the values
    If you try a value between, eg x=1 then you do not get more than 16
    So you do not want to be between and the only other alternative is outside
    So
    x<-4 or x>4
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    (Original post by ShivP_98)
    Hi,

    I'm a bit stuck on the procedure on finding the answer to x^2 > 16 and x^2 > 36 and similar. Can someone help me out? I understand that square routing gives + or - but stuck after that.

    Thanks
    I'd always make the coefficient of x^2 positive, factorise (or solve) the quadratic, draw a positive parabola with a horizontal line crossing it, label the roots of the quadratic and the answer will be obvious.
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    (Original post by ShivP_98)
    Hi,

    I'm a bit stuck on the procedure on finding the answer to x^2 > 16 and x^2 > 36 and similar. Can someone help me out? I understand that square routing gives + or - but stuck after that.

    Thanks
    For a quadratic inequality of this form, you can just use DOTS to factorise:

    x^2 &gt; 16 \Rightarrow x^2-16 &gt; 0 \Rightarrow (x+4)(x-4) &gt; 0

    Since ab &gt; 0 \Rightarrow either (a &gt; 0 \text{ and } b &gt; 0) \text{ or } (a &lt; 0 \text{ and } b &lt; 0) we must have either:

    x + 4 &gt; 0 \text{ and } x-4 &gt; 0 \Rightarrow x &gt;-4 \text{ and } x &gt; 4 \Rightarrow x &gt; 4

    or:

    x + 4 &lt; 0 \text{ and } x-4 &lt; 0 \Rightarrow x &lt; -4 \text{ and } x &lt; 4 \Rightarrow x &lt; -4

    i.e. if x^2 &gt; 16 then x is either less than -4 or greater than 4.

    You'll see this geometrically if you draw the graphs of y=x^2 and y=16 and look to see in which regions of the x-axis is the graph of the curve is above the graph of the line.
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    Still don't understand the graph thing - can anyone dumb it down a little?
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    From the mark scheme, the answer for something slightly different (x^2 < 16) is -4<x<4
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    (Original post by ShivP_98)
    From the mark scheme, the answer for something slightly different (x^2 < 16) is -4<x<4
    Of course it is ... Refer to my previous post
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    Ahh, I think I understand now. Thanks everyone for your help!
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    It looks like you may have written the question wrong in the OP. Your mark scheme answer is corrent for x^2 < 16

    However I've worked both of them out for you. Check below:

    x^2 > 16 (what you've asked for in the question)



    x^2 < 16 (what I think you we're meant to ask)

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    Thanks a lot, that helped also!
 
 
 
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