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Math Help - can anyone help with my AS level revision i have a few practise quests. but no ans.

  1. #1
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    can anyone help with my AS level revision i have a few practise quests. but no ans.

    i attmepted this one and got as far as possible but not sure where to go next
    find the solutions to x^2 -6x+2 leave answer as a surd i attempted it and ended up with 3+/- (sq. root of 7)
    and then it asks to then get the exact solution to the inequality x^2-6x+2 </= 0
    by substitution of (3+/- sq.root of 7) into x
    i got to a point where -5</= 0
    am i going right and does this mean X >/=5

    Thanks for help
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  2. #2
    Super Member craig's Avatar
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    You were right with your roots,

    x^2-6x+2 roots are, , however the equation in factorised form is, (x-3-\sqrt7)(x-3+\sqrt7).

    I presume that the inequatities are therefore;
    x\leq3+\sqrt7 and x\leq3-\sqrt7.

    It has been a while since I have done inequalities so I might be wrong?
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  3. #3
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    Quote Originally Posted by adeshSB View Post
    i attmepted this one and got as far as possible but not sure where to go next
    find the solutions to x^2 -6x+2 leave answer as a surd i attempted it and ended up with 3+/- (sq. root of 7)
    and then it asks to then get the exact solution to the inequality x^2-6x+2 </= 0
    by substitution of (3+/- sq.root of 7) into x
    i got to a point where -5</= 0
    am i going right and does this mean X >/=5

    Thanks for help
    Unfortunately, craig was not completely right. The factorization craig gives makes your inequality (x-3-\sqrt7)(x-3+\sqrt7)\le 0
    In order that a product of two numbers be negative, the numbers must have opposite sign. So either x-3-\sqrt{7}\le 0 and x-3+\sqrt{7}\ge 0 or [tex]x-3-\sqrt{7}\ge 0[\math] and x-3+\sqrt{7}\le 0. In the first case x\le 3+ \sqrt{7} and x\ge 3- \sqrt{7}: that is 3-\sqrt{7}\le x\le 3+ \sqrt{7}. In the second case, x\ge 3+ \sqrt{7} and x\le 3- \sqrt{7} which is impossible. All x such that 3- \sqrt{7}\le x\le 3+ \sqrt{7}.

    Another way to do this is to recognize that since x^2- 6x+ 2 is a continuous function so can change form negative to positive only where it is equal to 0. It is sufficient to check one value in each of the three intervals 3- \sqrt{7} and 3+\sqrt{7} break the real line into. 0< 3- \sqrt{7} and 0^2- 6(0)+ 2= 2> 0 so no x< 3- \sqrt{7} satisfies x^2- 6x+ 2< 0. 3-\sqrt{7}\le 3\le 3+\sqrt{7} and (3)^2- 6(3)+ 2= -7< 0 so every number between 3-\sqrt{7} and 3+ \sqrt{7} satisfies x^2- 6x+ 2< 0. Finally, 6> 3+ \sqrt{7} and (6)^2- 6(6)+ 2= 2> 0 so no number larger than 3+ \sqrt{7} satisfies the inequality.

    Finally, we can recognise that the graph of y= x^2- 6x+ 2 is a parabola opening upward. y will be less than 0 only for the portion of the parabola between the two roots 3- \sqrt{7} and 3+ \sqrt{7}.
    Last edited by mr fantastic; January 8th 2009 at 06:25 PM. Reason: Fixed some latex ... itex --> math and [math] --> [/math]
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  4. #4
    Super Member craig's Avatar
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    Sorry you are quite right had a feeling there was something missing when I posted
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