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Math Help - value of x coordinate

  1. #1
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    value of x coordinate

    hi, the question i've got is:
    find the exact value of the x coordinate at the point of intersection of-

    the line 2x + y = 7 and the curve y = 4x^2 - 8x - 5

    the answer is \frac{1}{4} (3 + \sqrt{57}) and \frac{1}{4}(3 - \sqrt{57})

    could someone show me the steps to arrive at this answer please?
    thanks
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  2. #2
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    Quote Originally Posted by mark View Post
    hi, the question i've got is:
    find the exact value of the x coordinate at the point of intersection of-

    the line 2x + y = 7 and the curve y = 4x^2 - 8x - 5

    the answer is \frac{1}{4} (3 + \sqrt{57}) and \frac{1}{4}(3 - \sqrt{57})

    could someone show me the steps to arrive at this answer please?
    thanks
    y=7-2x ---1

    y = 4x^2 - 8x - 5 --- 2

    when they intersect , 1=2 .

    7-2x=4x^2-8x-5

    4x^2-6x-12=0

    Use the quadratic formula to solve for x .
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  3. #3
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    yeah thats the problem, i did use the quadratic formula but came up with 3 \pm \sqrt57 but i don't understand where the \frac{1}{4} came from?
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  4. #4
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    Quote Originally Posted by mark View Post
    yeah thats the problem, i did use the quadratic formula but came up with 3 \pm \sqrt57 but i don't understand where the \frac{1}{4} came from?
    Oh ok ..

    x=\frac{-(-6)\pm\sqrt{36-4(4)(-12)}}{2(4)}

    =\frac{6\pm\sqrt{228}}{8}

    =\frac{6\pm2\sqrt{57}}{8}

    =\frac{3}{4}\pm\frac{\sqrt{57}}{4}

    so now what's common ?

     <br />
=\frac{1}{4}(3\pm\sqrt{57})<br />
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