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anonymous

  • one year ago

the values of x for which the graphs of y=x+2,y^2=4x intersect are: a. -2 and 2 b. -2 c. 2 d. 0 e. none of these

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  1. zepdrix
    • one year ago
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    \[\large\rm y=x+2,\qquad\qquad y^2=4x\]We have a couple ways to approach this... Hmm, let's try solving each equation for x.

  2. zepdrix
    • one year ago
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    In the first equation, subtract 2 gives us,\[\large\rm y-2=\color{orangered}{x}\]And in the second equation, dividing by 4 gives us,\[\large\rm \frac{y^2}{4}=\color{orangered}{x}\]They will intersect when these x's are the same, which means the equations are the same.\[\large\rm y-2=\frac{y^2}{4}\]

  3. zepdrix
    • one year ago
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    Multiply by 4,\[\large\rm 4y-8=y^2\]Move everything to one side,\[\large\rm y^2-4y+8=0\]

  4. zepdrix
    • one year ago
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    It looks like this `is not` going to factor nicely. Try using your quadratic formula:\[\large\rm y=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]Plug in the values, what does it tell you? :)

  5. anonymous
    • one year ago
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    so its b? @zepdrix

  6. zepdrix
    • one year ago
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    Hmm +_+ Well, we're trying to find our y-coordinate, and we can then use that to find the x-coordinate of intersection. What did you get for the y-value? Use the formula.

  7. anonymous
    • one year ago
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    -2 and 0 @zepdrix

  8. jiteshmeghwal9
    • one year ago
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    \(y=x+2\) \(y^2=4x\) From above eqn \(y^2=4(y-2)\) \(y^2-4y+8=0\) solve eqn either by quadratic formula or factorization.

  9. jiteshmeghwal9
    • one year ago
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    \[y=\frac{4 \pm \sqrt{16-32}}{2}\]and \[ x=y-2\]

  10. anonymous
    • one year ago
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    so im right @jiteshmeghwal9

  11. jiteshmeghwal9
    • one year ago
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    \[y=\frac{4 \pm 4 \iota}{2}=2\pm2 \iota\] \[x=y-2=2\pm2 \iota -2\]\[x= \pm 2 \iota\]

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