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    Well I'm glad you asked. I'm bored so I will demonstrate it to you, eager GCSE students.

    General quadratic:
    \displaystyle ax^2+bx+c=0 where a\neq 0

    Factor out the a and rearrange:
    \displaystyle a\left[ x^2+\frac{b}{a}x \right]+c=0

    \displaystyle \Rightarrow x^2+\frac{b}{a}x=-\frac{c}{a}

    Complete the square on the left:
    \displaystyle \left( x+\frac{b}{2a} \right)^2- \left( \frac{b}{2a} \right)^2=-\frac{c}{a}

    Move one term onto the right and tidy that place up (second step requires both fractions to be under the same denominator):
    \displaystyle \left( x+\frac{b}{2a} \right)^2=\frac{b^2}{4a^2}-\frac{c}{a}=\frac{b^2}{4a^2}-\frac{4ac}{4a^2}

    \displaystyle \Rightarrow \left( x+\frac{b}{2a} \right)^2=\frac{b^2-4ac}{4a^2}

    Square root both sides (the denominator square roots nicely but the numerator does not):
    \displaystyle x+\frac{b}{2a}=\pm \frac{\sqrt{b^2-4ac}}{2a}

    Get x on its own:
    \displaystyle x=-\frac{b}{2a} \pm \frac{\sqrt{b^2-4ac}}{2a}

    The denominator is the same so we can combine the fractions very easily:
    \displaystyle x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

    and there you have it! You're welcome. Feel free to ask about anything you don't understand.
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    (Original post by RDKGames;[url="tel:66845224")
    66845224[/url]]Well I'm glad you asked. I'm bored so I will demonstrate it to you, eager GCSE students.

    General quadratic:
    ax^2+bx+c=0

    Factor out the a and rearrange:
    a[x^2+\frac{b}{a}x]+c=0

    \rightarrow x^2+\frac{b}{a}x=-\frac{c}{a}

    Complete the square on the left:
    (x+\frac{b}{2a})^2-(\frac{b}{2a})^2=-\frac{c}{a}

    Move one term onto the right and tidy that place up (second step requires both fractions to be under the same denominator):
    (x+\frac{b}{2a})^2=\frac{b^2}{4a  ^2}-\frac{c}{a}
    \rightarrow (x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2}

    Square root both sides (the denominator square roots nicely but the numerator does not):
    x+\frac{b}{2a}=\pm \frac{\sqrt{b^2-4ac}}{2a}

    Get x on its own:
    x=-\frac{b}{2a} \pm \frac{\sqrt{b^2-4ac}}{2a}

    The denominator is the same so we can add them:
    x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

    and there you have it! You're welcome.
    I actually enjoy reading and learning about stuff like this 😂😂😂
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    (Original post by danielwinstanley)
    I actually enjoy reading and learning about stuff like this 😂😂😂
    Good! The more you know the better
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    shouldn't +c be over a or in brackets to be multiplied by a on the second line?
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    wait doesnt matter, didnt fully read through yet
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    (Original post by TSRPAV)
    shouldnt +c be over a on the second line?
    Could indeed be but I didn't factor a out of everything so to speak. I could've but I didn't. It checks out in line 3 as I move c across and divide both sides by a. Same result as if I had c/a in line 2.
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    Try the cubic formula.
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    (Original post by B_9710)
    Try the cubic formula.
    I'm alright, thanks for the offer though.
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    (Original post by RDKGames)
    I'm alright, thanks for the offer though.
    It took me 5 pages. The difference of a single power of x ay.
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    (Original post by B_9710)
    It took me 5 pages. The difference of a single power of x ay.
    Now get a general result for any polynomial of nth degree.
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    (Original post by RDKGames)
    Now get a general result for any polynomial of nth degree.
    Are you aware that this is impossible?
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    (Original post by B_9710)
    Are you aware that this is impossible?
    Yes.
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    (Original post by RDKGames)
    I'm alright, thanks for the offer though.
    How would you solve this equation.
    (x^2+2x+1)^2 -5(x^2+2x+1)+6=0. I tried expanding it all out it I couldn't do anything from there.
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    (Original post by Ano9901whichone)
    How would you solve this equation.
    (x^2+2x+1)^2 -5(x^2+2x+1)+6=0. I tried expanding it all out it I couldn't do anything from there.
    don't expand it out! That will take far more time than necessary.

    If you look at this example it may help:

    x^4 - 13x^2 + 36 = 0
    Now, at first it may not seem like it, but this is a quadratic 'in disguise'.
    You want to get this equation into the form of a quadratic (ax^2 + bx +c = 0) as you know how to solve quadratics.
    To do this, you want the first term to be to the power of 2, the second to the power of 1, and the last to be to the power of 0.

    (x^2)^2 - 13(x^2)^1 + 36(x^2)^0

    Looking at the rules of indices, you should see that this equation is equivalent to the original.

    Now this looks complicated, so you can sub in a letter that would replace x^2 in your equation to clean it up a little.
    For instance, let y = x^2, and sub y into your new equation.

    y^2 - 13x + 36 = 0

    And now, this is a quadratic, something you know how to solve.
    Go ahead and factorise this:

    (y - 4)(y - 9) = 0

    And solving this, you get y = 4 or y = 9.
    BUT, this isn't your answer. If you remember from earlier, we said that y = x^2. So to work out what x is, you need to get the square roots of your answers, which will be +-2 and +-3

    I know that may have been a long example, but can you see how that applies to your question?
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    (Original post by Ano9901whichone)
    How would you solve this equation.
    (x^2+2x+1)^2 -5(x^2+2x+1)+6=0. I tried expanding it all out it I couldn't do anything from there.
    Expanding is a pain and will lead you to quartics which you cannot solve at GCSE. You need to use a trick of substitution.

    Let y=x^2+2x+1 and substitute it in to get y^2-5y+6=0 which you can easily solve for y

     \rightarrow (y-3)(y-2)=0

    Therefore y=3 and y=2

    Now replace the y with the quadratic we defined it to be and get yourself

    x^2+2x+1=3

    and

    x^2+2x+1=2

    then proceed to solve them both for x.
 
 
 
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