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Math Help - Quadratic Equation Part 1

  1. #1
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    Quadratic Equation Part 1

    How to do these?

    1. Find the real roots of 2x^2+3x-1=0. Give your answer in surd form.

    2. Find the range of values of q if x^2+(q+1)x+q+1=0 has complex roots.

    3. Find the range of values of m if x^2+x+1=m(x+2) has real and distinct roots.

    4. Find the range of values of k if x^2+(k-3)x+k=0 has roots with same sign.

    Answer:
    1. \frac{1}{2}(-3 \pm \sqrt{17})

    2. -1<q<3

    3. m<-3-2\sqrt{3},m>-3+2\sqrt{3}

    4. k>0
    Last edited by cloud5; June 24th 2009 at 06:41 AM.
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  2. #2
    Senior Member TheAbstractionist's Avatar
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    For (4), you got part of it correct (namely the product of the roots is positive). But you also want your roots to be real, since it doesn’t make any sense to talk about two conjugate complex numbers having the same sign.
    Last edited by TheAbstractionist; June 24th 2009 at 07:40 AM.
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  3. #3
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    1. Find the real roots of 2x^2+3x-1=0. Give your answer in surd form.
    Use the quadratic formula:
    \begin{aligned}<br />
x &= \frac{-3 \pm \sqrt{3^2 - 4(2)(-1)}}{2(2)} \\<br />
&= \frac{-3 \pm \sqrt{9 + 8}}{4} \\<br />
&= \frac{-3 \pm \sqrt{17}}{4} \\<br />
&= \frac{1}{4}(-3 \pm \sqrt{17})<br />
\end{aligned}
    Hmm, I'm getting a different answer.

    2. Find the range of values of q if x^2+(q+1)x+q+1=0 has complex roots.
    A quadratic equation in the form ax^2 + bx + c = 0 has complex roots if the discriminant b^2 - 4ac < 0. So
    a = 1
    b = q + 1
    c = q + 1
    \begin{aligned}<br />
b^2 - 4ac &< 0 \\<br />
(q + 1)^2 - 4(1)(q + 1) &< 0 \\<br />
q^2 + 2q + 1 - 4q - 4 &< 0 \\<br />
q^2 - 2q - 3 &< 0 \\<br />
(q - 3)(q + 1) &< 0<br />
\end{aligned}

    Make a sign chart. Draw a number line and the critical points -1 and 3 (you get these by setting each factor equal to 0).
    Code:
    ----+----+----+----+----+----+----+----
                 -1                   3
    You have 3 intervals to test:
    (-∞, -1)
    (-1, 3)
    (3, ∞)

    From each interval, pick a number inside it to test into the inequality (q - 3)(q + 1) < 0, in order to determine the sign. The signs of the polynomial for each interval is as follows:
    Code:
    ----+----+----+----+----+----+----+----
          pos    -1        neg        3 pos
    So the answer is -1 < q < 3.


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  4. #4
    Member SENTINEL4's Avatar
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    Quote Originally Posted by yeongil View Post
    Use the quadratic formula:
    \begin{aligned}<br />
x &= \frac{-3 \pm \sqrt{3^2 - 4(2)(-1)}}{2(2)} \\<br />
&= \frac{-3 \pm \sqrt{9 + 8}}{4} \\<br />
&= \frac{-3 \pm \sqrt{17}}{4} \\<br />
&= \frac{1}{4}(-3 \pm \sqrt{17})<br />
\end{aligned}
    Hmm, I'm getting a different answer.
    I find this too...
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  5. #5
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    3. Find the range of values of m if x^2+x+1=m(x+2) has real and distinct roots.
    This is the same as #2, but this time, since we want real and distinct roots, the discriminant has to be positive, or b^2 - 4ac > 0. So
    \begin{aligned}<br />
x^2 + x + 1 &= m(x + 2) \\<br />
x^2 + x + 1 &= mx + 2m \\<br />
x^2 + x - mx + 1 - 2m &= 0 \\<br />
x^2 + (1 - m)x + (1 - 2m) &= 0 \\<br />
 \end{aligned}

    a = 1
    b = 1 - m
    c = 1 - 2m
    \begin{aligned}<br />
b^2 - 4ac &> 0 \\<br />
(1 - m)^2 - 4(1)(1 - 2m) &> 0 \\<br />
1 - 2m + m^2 - 4 + 8m &> 0 \\<br />
m^2 + 6m - 3 &> 0<br />
\end{aligned}

    I want to find the zeros of the polynomial m^2 + 6m - 3. Let's pretend for a moment that it equals 0 and solve for m:
    \begin{aligned}<br />
m^2 + 6m - 3 &= 0 \\<br />
m^2 + 6m &= 3 \\<br />
m^2 + 6m + 9 &= 3 + 9 \\<br />
(m + 3)^2 &= 12 \\<br />
m + 3 &= \pm\sqrt{12} \\<br />
m &= -3 \pm 2\sqrt{3} \\<br />
\end{aligned}

    Make a sign chart. Draw a number line and the critical points (the zeros we found earlier):
    Code:
    ---------+-------------------+---------
           -6.46                0.46
    ( 0.46 \approx -3 + 2\sqrt{3} and -6.46 \approx -3 - 2\sqrt{3})

    You have 3 intervals to test:
    (-\infty,\;-3 - 2\sqrt{3})
    (-3 - 2\sqrt{3},\;-3 + 2\sqrt{3})
    (-3 + 2\sqrt{3}, \infty)

    From each interval, pick a number inside it to test into the inequality m^2 + 6m - 3 > 0, in order to determine the sign. The signs of the polynomial for each interval is as follows:
    Code:
    ---------+-------------------+---------
       pos -6.46      neg       0.46 pos
    So the answer is m<-3-2\sqrt{3},m>-3+2\sqrt{3}.


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  6. #6
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    Quote Originally Posted by yeongil View Post
    Use the quadratic formula:
    \begin{aligned}<br />
x &= \frac{-3 \pm \sqrt{3^2 - 4(2)(-1)}}{2(2)} \\<br />
&= \frac{-3 \pm \sqrt{9 + 8}}{4} \\<br />
&= \frac{-3 \pm \sqrt{17}}{4} \\<br />
&= \frac{1}{4}(-3 \pm \sqrt{17})<br />
\end{aligned}
    Hmm, I'm getting a different answer.
    ! I got that answer too! But the answer in my book is \frac{1}{2}... Maybe there is a mistake in my book?
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  7. #7
    Senior Member Stroodle's Avatar
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    Yeah. Looks like a mistake. I got \frac{1}{4}(-3\pm\sqrt{17}) too
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