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    So to reduce making threads like every night, im going to concentrate every C1 related issues i have on this thread

    Can anyone check my answer for this inequalities question?

    find the set value for
    a) 1 - 2x < 4
    b) x^2 - 6x + 8 ≥ 0


    for a) my answer is x > 3/2
    for b) my answer is 2 ≥ x ≥ 4
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    (Original post by ihatePE)
    for a) my answer is x > 3/2
    for b) my answer is 2 ≥ x ≥ 4
    a.) Nope
    b.) Nope
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    (Original post by ihatePE)
    So to reduce making threads like every night, im going to concentrate every C1 related issues i have on this thread

    Can anyone check my answer for this inequalities question?

    find the set value for
    a) 1 - 2x < 4
    b) x^2 - 6x + 8 ≥ 0


    for a) my answer is x > 3/2
    for b) my answer is 2 ≥ x ≥ 4
    For a) it can be done in the same way you rearrange an equation, you forgot a minus sign.

    For b) you need to consider the critical values and you find them by setting x^2-6x+8 = 0 and then draw the curve and find the regions where
    x^2-6x+8 \geq 0 . The region you have is where x^2-6x+8 \leq 0
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    Here's a sketch of the second graph and what you're after (where the parabola is above the x axis).
    Name:  image.jpg
Views: 420
Size:  37.7 KB
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    Both incorrect.
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    It's always worth doing a quick sketch of the graph or a 'sign diagram' for inequalities of polynomials (or functions like 1/x) with degree greater than 1.
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    (Original post by B_9710)
    Here's a sketch of the second graph and what you're after (where the parabola is above the x axis).
    Awh come on, you're practically giving away the answer. No fun.
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    (Original post by NotNotBatman)
    For a) it can be done in the same way you rearrange an equation, you forgot a minus sign.

    For b) you need to consider the critical values and you find them by setting x^2-6x+8 = 0 and then draw the curve and find the regions where
    x^2-6x+8 \geq 0 . The region you have is where x^2-6x+8 \leq 0
    ok i got a) -3/2

    but for b) im still unsure, x^2 - 6x + 8 ≥ 0
    which ive factorised to (x-2)(x-4)
    so x=2 or x=4 and i sketched the graph.
    looking back at the original equation y ≥ 0 so im looking at the positive y quadrant which gave me 2≥x≥4

    i still dont know where i went wrong there
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    (Original post by ihatePE)
    ok i got a) -3/2

    but for b) im still unsure, x^2 - 6x + 8 ≥ 0
    which ive factorised to (x-2)(x-4)
    so x=2 or x=4 and i sketched the graph.
    looking back at the original equation y ≥ 0 so im looking at the positive y quadrant which gave me 2≥x≥4

    i still dont know where i went wrong there
    Which quadrant you look at has nothing to do with the inequality. Look at the graph you sketched and look at what points the parabola is GREATER than 0, and look what values of y YOUR inequality satisfies. You should see your error.
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    (Original post by B_9710)
    Here's a sketch of the second graph and what you're after (where the parabola is above the x axis).
    Name:  image.jpg
Views: 420
Size:  37.7 KB
    that's helpful but why is the white bit shaded white?
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    (Original post by ihatePE)
    ok i got a) -3/2

    but for b) im still unsure, x^2 - 6x + 8 ≥ 0
    which ive factorised to (x-2)(x-4)
    so x=2 or x=4 and i sketched the graph.
    looking back at the original equation y ≥ 0 so im looking at the positive y quadrant which gave me 2≥x≥4

    i still dont know where i went wrong there
    y\geq 0 is "above the x axis", so for which values of x is the curve above the x axis? Look at B_9710's post and try more questions to make sure you've got the idea.
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    (Original post by NotNotBatman)
    y\geq 0 is "above the x axis", so for which values of x is the curve above the x axis? Look at B_9710's post and try more questions to make sure you've got the idea.
    so 2≥x or x≥4??? my brain isnt working so well at night
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    (Original post by ihatePE)
    so 2≥x or x≥4??? my brain isnt working so well at night
    Yes.
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    (Original post by NotNotBatman)
    Yes.
    omg omg...so the question is ''find solution'' and then theres a question ''find the range'' are they different? is that where i got it mixed up?
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    (Original post by ihatePE)
    omg omg...so the question is ''find solution'' and then theres a question ''find the range'' are they different? is that where i got it mixed up?
    You find the solution to an equation, that is when there is an '=', these solutions give us the roots (where the graph crosses the x axis). These are critical values which are then used to find the range; that's when you have an inequality.
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    (Original post by ihatePE)
    that's helpful but why is the white bit shaded white?
    The curve you gave us below the x axis in the white region (i.e.  y\leq 0 ). Either side of the roots, in the orange regions the curve is above the x axis (i.e.  y\geq 0 ). The orange regions are the one you want, meaning it is the region where the curve is greater or equal to 0 - which can be clearly seen as the curve is above the x axis.
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    (Original post by B_9710)
    The curve you gave us below the x axis in the white region (i.e.  y\leq 0 ). Either side of the roots, in the orange regions the curve is above the x axis (i.e.  y\geq 0 ). The orange regions are the one you want, meaning it is the region where the curve is greater or equal to 0 - which can be clearly seen as the curve is above the x axis.
    Thanks I got a better idea of this topic now and will do more questions tomorrow

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    (Original post by RDKGames)
    Awh come on, you're practically giving away the answer. No fun.
    I think it's always good to encourage people to think graphically. Sometimes people get so caught up in the algebra, they forget what it all means in terms of the graphs.
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    Highly recommend Jack Brown on youtube, he has a playlist for C1,C2,C3,C4 and makes everything so easy to understand
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    I have terrible handwriting but made a worked solution.
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