math

Here's my question: Determine if each system has one, two, or no solutions. (Show your work - 3 marks each)

a)y=2x^2-2x+1 and y=3x-5

b)y=x^2+3x-16 and y=-x^2-8x-18

Here's my work on the question:

a)

x=(-b plus/minus sign (square root b^2-4ac)/(2a)
x=(--2 plus/minus sign (square root -2^2-4(2)(1)/(2(2)
x=(2 plus/minus sign (square root -4-8)/(4)
x=( 2 plus/minus sign (non-real answer)/(4)

x=(-b plus/minus sign (square root b^2-4ac)/(2a)
x=(--5 plus/minus sign (square root -5^2-4(3)/2(3)
x=(5 plus/minus sign(square root -25-12)/6
x=(5 plus/minus sign(square root -37)/6
x=(5 plus/minus sign (non-real answer)/6


b)
x=(-b plus/minus sign (square root b^2-4ac)/(2a)
x=(-3 plus/minus sign (square root 3^2-4(1)(-16)/(2(1))
x=(-3 plus/minus sign (square root 9--64)/2
x=(-3 plus/minus sign (square root 73)/2
x=(-3 plus/minus sign (8.54)/(2)
x=4.27

x=(-b plus/minus sign (square root b^2-4ac)/(2a)
x=(--8 plus/minus sign(square root -8^2-4(-1)(-18)/(2(-1)
x=(8 plus/minus sign(square root-64-72)/(-2)
x=(8 plus/minus sign(square root -136)/(-2)
x=(8 plus/minus sign(non-real answer)/(-2)

Can you please help me where I went wrong and help correct my errors!

asked by Alex
  1. first, do a google on special characters to find the ±√ symbols, which you can then copy and paste. All those words are just too ugly for words! You have

    y=2x^2-2x+1 and y=3x-5, so set them equal:

    2x^2-2x+1 = 3x-5
    2x^2-5x+6 = 0

    The discriminant (b^2-4ac) is negative, so there are no solutions.

    For the other,

    y=x^2+3x-16 and y=-x^2-8x-18

    x^2+3x-16 = -x^2-8x-18
    2x^2+11x+2 = 0

    the discriminant is positive, so it has two distinct solutions.

    See the graphs at

    http://www.wolframalpha.com/input/?i=plot+y%3D2x^2-2x%2B1+%2C+y%3D+3x-5

    http://www.wolframalpha.com/input/?i=plot+y%3Dx^2%2B3x-16+%2C+y%3D-x^2-8x-18

    posted by Steve

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