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2x^2+(3-K)x+K+3=0 Maths help? Watch

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    I have to find the range of values K could have for 2x^2+(3-K)x+K+3=0 to have two real distinct roots. If someone understands this, please help out!
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    (Original post by ~xSarahx~)
    I have to find the range of values K could have for 2x^2+(3-K)x+K+3=0 to have two real distinct roots. If someone understands this, please help out!
    for the quadratic  \displaystyle ax^2 +bx + c to have two distinct roots  \displaystyle b^2 - 4ac > 0
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    (Original post by DylanJ42)
    for the quadratic  \displaystyle ax^2 +bx + c to have two distinct roots  \displaystyle b^2 - 4ac > 0
    Thanks! I understand that but I just honestly don't understand how to figure this whole thing out :/ I can't really find any good examples online and it's just really confusing me.
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    (Original post by ~xSarahx~)
    Thanks! I understand that but I just honestly don't understand how to figure this whole thing out :/ I can't really find any good examples online and it's just really confusing me.
    In your quadratic what do you think a=, b= and c= ?


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    (Original post by ~xSarahx~)
    Thanks! I understand that but I just honestly don't understand how to figure this whole thing out :/ I can't really find any good examples online and it's just really confusing me.
    do you understand the quadratic formula, or at least how it gets you the roots of a quadratic?

    well the quadratic formula tells us that the roots of the quadratic equation  \displaystyle ax^2 + bx + c = 0 are  \displaystyle x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

    As you know you cannot square root a negative number so in order for real roots to exist the part under the square root sign must be > 0.

    If the part under the square root is zero we end up with  \displaystyle x = \frac{-b \pm \sqrt{0}}{2a} which means both roots are the same. You still get 2 roots but they are same ie not distinct roots.

    Finally if the part under the sqaure root is positive then you get two distinct roots, namely  \displaystyle x = \frac{-b + \sqrt{b^2 - 4ac}}{2a} and  \displaystyle x = \frac{-b - \sqrt{b^2 - 4ac}}{2a}

    In Summary;

    So if a quadratic has no real roots then the part under the square root sign is negative ie  \displaystyle b^2 - 4ac < 0

    If a quadratic has a repeated root (this means both x values you find from the quadratic formula are the same) then the part under the square root sign is equal to 0 ie  \displaystyle b^2 - 4ac = 0

    Finally if a quadratic has two distinct roots then the part under the square root is > 0 ie  \displaystyle b^2 - 4ac > 0

    I hope this clears it up at least a little for you
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    (Original post by gdunne42)
    In your quadratic what do you think a=, b= and c= ?


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    That's another issue I'm having.
    so
    a= 2xˆ2

    but I'm not sure where that x between (3-k) and k+3 goes.

    is it
    b= (3-k)x OR b=(3-k)

    c=k+3 OR c=x+K+3

    I've never done this with k before which is why i'm so useless at this...
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    (Original post by DylanJ42)
    do you understand the quadratic formula, or at least how it gets you the roots of a quadratic?

    well the quadratic formula tells us that the roots of the quadratic equation  \displaystyle ax^2 + bx + c = 0 are  \displaystyle x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

    As you know you cannot square root a negative number so in order for real roots to exist the part under the square root sign must be > 0.

    If the part under the square root is zero we end up with  \displaystyle x = \frac{-b \pm \sqrt{0}}{2a} which means both roots are the same. You still get 2 roots but they are same ie not distinct roots.

    Finally if the part under the sqaure root is positive then you get two distinct roots, namely  \displaystyle x = \frac{-b + \sqrt{b^2 - 4ac}}{2a} and  \displaystyle x = \frac{-b - \sqrt{b^2 - 4ac}}{2a}

    In Summary;

    So if a quadratic has no real roots then the part under the square root sign is negative ie  \displaystyle b^2 - 4ac < 0

    If a quadratic has a repeated root (this means both x values you find from the quadratic formula are the same) then the part under the square root sign is equal to 0 ie  \displaystyle b^2 - 4ac = 0

    Finally if a quadratic has two distinct roots then the part under the square root is > 0 ie  \displaystyle b^2 - 4ac > 0

    I hope this clears it up at least a little for you
    Thank you for explaining it out like that! The question is starting to make more sense for me!
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    Perhaps this way will make it clearer.

    For the equation (a)x^2 + (b)x + (c) = 0, we want b^2-4ac > 0.

    Your equation is

    (2)x^2 + (3-k)x + (k+3) = 0

    Perhaps looking at it that way will make it a little easier to understand.
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    (Original post by ~xSarahx~)
    That's another issue I'm having.
    so
    a= 2xˆ2

    but I'm not sure where that x between (3-k) and k+3 goes.

    is it
    b= (3-k)x OR b=(3-k)

    c=k+3 OR c=x+K+3

    I've never done this with k before which is why i'm so useless at this...

    a is the coefficient of the x^2 term = 2
    b is the coefficient of the x term = (3-k)
    c is the constant at the end = (k+3)


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    (Original post by ~xSarahx~)
    Thank you for explaining it out like that! The question is starting to make more sense for me!
    youll get by if you learn off the summary part at the bottom. Just know whether to use  \displaystyle b^2 - 4ac > 0 , \displaystyle  b^2 - 4ac = 0 ,  \displaystyle  b^2 - 4ac < 0 depending on whether they ask for 2 distinct roots, 1 repeated root or 0 real roots
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    (Original post by sammygmfc)
    Perhaps this way will make it clearer.

    For the equation (a)x^2 + (b)x + (c) = 0, we want b^2-4ac > 0.

    Your equation is

    (2)x^2 + (3-k)x + (k+3) = 0

    Perhaps looking at it that way will make it a little easier to understand.
    Oh oh ohhhh!

    So...perhaps

    (3-k)ˆ2 - (4x2x[k+3]) > 0 would be..right? Or am I still not getting it?
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    (Original post by ~xSarahx~)
    Oh oh ohhhh!

    So...perhaps

    (3-k)ˆ2 - (4x2x[k+3]) > 0 would be..right? Or am I still not getting it?
    That's it


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    (Original post by ~xSarahx~)
    Oh oh ohhhh!

    So...perhaps

    (3-k)ˆ2 - (4x2x[k+3]) > 0 would be..right? Or am I still not getting it?
    That's right. Then solve to find the set of values for which k could be.
 
 
 
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