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xKingx

  • 3 years ago

Can someone show me how I would solve this using the Quadratic Formula? 2x^2 – 16x + 32 = 0

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  1. haleylou56
    • 3 years ago
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    do you know the quadratic form?

  2. anas2000
    • 3 years ago
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    I think x=4

  3. hartnn
    • 3 years ago
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    Compare your quadratic equation with \(ax^2+bx+c=0\) find a,b,c then the two roots of x are: \(\huge{x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}\)

  4. xKingx
    • 3 years ago
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    The +- part always confuses me. Could you demonstrate how I'd substitute an equation like this into the form?

  5. AccessDenied
    • 3 years ago
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    \[ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \text{ means:} \\ \quad x_1 = \frac{-b + \sqrt{b^2 - 4ac}}{2a} \text{, and} \\ \quad x_2 = \frac{-b - \sqrt{b^2 - 4ac}}{2a} \text{ are the two solutions} \] Basically, its a more compact way of writing both solutions. The typical method is to substitute your values, simplify your sqrt(b^2 - 4ac) and -b, and then deal with splitting the \(\pm\) into two solutions + and -.

  6. pkjha3105
    • 3 years ago
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    2x^2-8x-8x+32=0 2x(x-4)-8(x-4)=0 (2x-8)(x-4)=0 x=4,2

  7. xKingx
    • 3 years ago
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    Can I ask why you wrote it as 2x^2-8x-8x+32=0

  8. xKingx
    • 3 years ago
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    its supposed to be b+-4ac?

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