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how to factorise without calculator watch

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    How would you factorise 1-3x-2rootx = 0 without a calculator?
    Thanks
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    let  \sqrt{x} = y and solve for y (since u would get a quadratic equation in terms of y) and when you've got the solutions for y, resub in  y = \sqrt{x}
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    (Original post by RickHendricks)
    let  \sqrt{x} = y and solve for y (since u would get a quadratic equation in terms of y) and when you've got the solutions for y, resub in  y = \sqrt{x}
    Oh, is this a "hidden quadratic"?
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    (Original post by boredsatan)
    Oh, is this a "hidden quadratic"?
    In a way, yes, but remember to resub in the y values.
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    (Original post by RickHendricks)
    In a way, yes, but remember to resub in the y values.
    How do we generally recognise hidden quadratics?
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    (Original post by boredsatan)
    How do we generally recognise hidden quadratics?
    in general it depends on the coefficients. IF the coffieficents are squares of each other it means it could be a polynomial.
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    (Original post by boredsatan)
    How do we generally recognise hidden quadratics?
    hidden or disguised quads are fairly easy to spot.
    If we can rearrange the equation to get

    ax^2n + bx^n + c = 0

    then it will be a disguised quad.

    In other words, the biggest power is double the smaller power.

    To solve, the most straightforward way, to avoid some silly mistakes, is to sub y = x^n, and then y-squared = x^2n and it'll be a more recognisable, less cumbersome appearance.
    But as mentioned above, be sure to sub back in and solve for the original variable (in the case of my example, 'x' )
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    (Original post by begbie68)
    hidden or disguised quads are fairly easy to spot.
    If we can rearrange the equation to get

    ax^2n + bx^n + c = 0

    then it will be a disguised quad.

    In other words, the biggest power is double the smaller power.

    To solve, the most straightforward way, to avoid some silly mistakes, is to sub y = x^n, and then y-squared = x^2n and it'll be a more recognisable, less cumbersome appearance.
    But as mentioned above, be sure to sub back in and solve for the original variable (in the case of my example, 'x' )
    More generally they have powers in arithmetic progression,

    for example, x^7-9x^4+8x=0.
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    The line y = ax-1 is is a tangent to the curve y = x^(1/2) + d at the point (9,c) where a, c and d are real constants. Find a, c and d?

    y = x^(1/2) + d dy/dx = 1/(2rootx) when x = 9, dy/dx = 1/(2root9) = 1/6 so the gradient of the tangent is 1/6 therefore a = 1/6 so m = 1/6, (x1,y1) = (9,c) y - c = (1/6)(x-9) y - c = x/6 - 9/6 y = x/6 - 9/6+c -9/6+c = -1 therefore c = 1/2 so we have the point (9,1/2) 9^(1/2) + d = 1/2 d+3 = 1/2 therefore d = -5/2 so our values are a = 1/6, c = 1/2, d = -5/2 can someone tell me if i'm correct? Thanks
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    (Original post by boredsatan)
    The line y = ax-1 is is a tangent to the curve y = x^(1/2) + d at the point (9,c) where a, c and d are real constants. Find a, c and d?

    y = x^(1/2) + d dy/dx = 1/(2rootx) when x = 9, dy/dx = 1/(2root9) = 1/6 so the gradient of the tangent is 1/6 therefore a = 1/6 so m = 1/6, (x1,y1) = (9,c) y - c = (1/6)(x-9) y - c = x/6 - 9/6 y = x/6 - 9/6+c -9/6+c = -1 therefore c = 1/2 so we have the point (9,1/2) 9^(1/2) + d = 1/2 d+3 = 1/2 therefore d = -5/2 so our values are a = 1/6, c = 1/2, d = -5/2 can someone tell me if i'm correct? Thanks
    Sure, but that calculation for c looks messy.

    a = \dfrac{1}{6} as you said... so y = \dfrac{1}{6}x - 1 by substituting it into the line's equation.

    Since the point (9,c) is on the tangent, we have c = \dfrac{1}{6}(9) - 1 = \dfrac{1}{2}.


    Please, do post a new question on a new thread though.
 
 
 
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