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Quadratics

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Size:  505.1 KBHow do you do question 7, 8, 9, 10 and 11? I know you're meant to show working but I don't have a clue what to do
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    (Original post by Lucofthewoods)
    How do you do question 7, 8, 9, 10 and 11? I know you're meant to show working but I don't have a clue what to do
    Key Theory Required:
    For a quadratic of the form ax^{2}+bx+c
    if b^{2}-4ac>0 there are two distinct solutions.
    if b^{2}-4ac=0 there is a repeated root.
    if b^{2}-4ac<0 there are no real solutions.

    I will show the method for question 7 then it should be a variant on the method for the rest of the questions.

    Question 7:
    x^{2}+6x>3x-4
    x^{2}+3x+4>0
    b^{2}-4ac=3^{2}-4\times 1 \times 4 = -7 < 0
    Therefore there are no real roots to the quadratic.
    When x=0, y=4 and given that there are no real roots,
     x^{2}+6x>3x-4 for  x \in \mathbb{R}

    Hints for other questions in spoiler:
    Spoiler:
    Show
    Q8: Same method as Q7.
    Q9: Same method as Q7 but you are proving the opposite (can be done by contradiction).
    Q10: Hint: b^{2}-4ac=0
    Q11: Hint: b^{2}-4ac=0
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    you need to examine the discriminant as the previous poster said.
 
 
 
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