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    Can someone check to see if I am doing this right?

    Factorise and Solve the Inequality:

    4x^2 < 25

    4x^2 - 25=0

    (2x - 5)^2

    (x - 5/2)=0

    X= 5/2

    -5/2 < x < 5/2

    I am self teaching Maths, I have the questions and answers, I know the answer is correct, I would however feel at ease knowing my workings are correct.
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    (Original post by Shakaa)
    Can someone check to see if I am doing this right?

    Factorise and Solve the Inequality:

    4x^2 < 25

    4x^2 - 25=0

    (2x - 5)^2


    (x - 5/2)=0

    X= 5/2

    -5/2 < x < 5/2

    I am self teaching Maths, I have the questions and answers, I know the answer is correct, I would however feel at ease knowing my workings are correct.
    You sure this part is correct?

    (2x - 5)^2 \not=4x^2 - 25
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    You could get to the answer in just two steps:

    4x^2 < 25
    x^2 < 25/4
    |x| < 5/2 (|x| means the magnitude i.e. size of x)

    Also a^2-b^2=(a-b)(a+b) for when you have the difference of two squares.
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    Doing it with your method, OP, it would be

    4x²<25
    <=> 4x²-25<0
    <=> (2x-5)(2x+5)< 0

    Now either you do the
    x -inf - 5/2 5/2 +inf

    2x-5 Negative | neg | pos
    2x+5 Negative | pos | pos

    4x²-25 Positive | neg | pos

    Therefore 4x²-25<0 <=> x belongs to ]-5/2,5/2[

    Either you know that for a polynomial P(x)=ax²+bx+c,
    the polynomial is of the sign of a outside of the "roots", and of the contrary sign inside the roots

    Here, a=4>0 so P(x)>0 for x belonging to ]-inf, -5/2[ u ]5/2,+inf[
    and P(x)<0 for x E ]-5/2,5/2[
 
 
 

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