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For what values of $C$ will $2x^2 + 7x + C = 0$ have? below are a few choices of the value $C$:

$a)$ 2 answers

$b)$ 1 answer

$c)$ 0 answers

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  • $\begingroup$ Are you familiar with the quadratic formula? $\endgroup$ – Nick Mar 4 '14 at 23:25
  • $\begingroup$ Could you post what you've tried so far? This will help others know where you're getting stuck and help them give you better answers. $\endgroup$ – recursive recursion Mar 4 '14 at 23:29
  • $\begingroup$ yeap thanks i get it already $\endgroup$ – carry Mar 5 '14 at 8:46
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The discriminant of the above quadratic polynomial is equal to $$\Delta = 7^2-4(2)C=49-8C$$

  1. First Case: If $\Delta<0$, that is if $$49-8C<0 \Leftrightarrow C>\frac{49}{8}$$ then the polynomial has no (real) solutions.
  2. Second Case: If $\Delta=0$, that is if $$49-8C=0 \Leftrightarrow C=\frac{49}{8}$$ then the polynomial has one (double) solution $x_1$, which is given by $$x_1=\frac{-7\pm\sqrt{\Delta}}{2(2)}=\frac{-7\pm\sqrt{0}}{4}=\frac{-7\pm0}{4}=-\frac{7}{4}$$
  3. Third Case: If $\Delta>0$, that is if $$49-8C>0 \Leftrightarrow C>\frac{49}{8}$$ then the polynomial has two (distinct) solution $x_{1,2}$, which are given by $$x_{1,2}=\frac{-7\pm\sqrt{\Delta}}{2(2)}=\frac{-7\pm\sqrt{49-8C}}{4}$$ Now the two solutions depend on the value of $C$ and cannot be computed unless this value is known.
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  • $\begingroup$ oh so we are not suppose to find the exact number of C. Thanks, i get it $\endgroup$ – carry Mar 5 '14 at 0:00
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Think about the formula for the solution to the quadratic equation. It gives two answers. Thus this equation has one root if and only if those two answers are the same.

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  • $\begingroup$ what? what do you mean? can you give examples? $\endgroup$ – carry Mar 4 '14 at 23:37
  • $\begingroup$ What is the formula for the solutions to $ax^2+bx+c=0$? $\endgroup$ – Stella Biderman Mar 4 '14 at 23:39
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Hint: use the discriminant to find the number of roots.

http://en.wikipedia.org/wiki/Discriminant#Nature_of_the_roots

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  • $\begingroup$ but there is still two unknown~ square root of 7^2-4(2)C $\endgroup$ – carry Mar 4 '14 at 23:27
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The quadratic formula is one of the most useful in pre-calculus. It is of the form $$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a},$$

for an equation of the form $ax^2+bx+c=0$. The discriminant, $b^2-4ac$ gives valuable information about the equation. If $b^2-4ac\gt0$ then the quadratic equation has two solutions, if $b^2-4ac=0$ then the quadratic equation has one solution, and if $b^2-4ac\lt0$ then the quadratic equation has no solutions (technically no real solutions). Thus, if you have an equation of the form $$2x^2+7x+C=0$$ then you use the discriminant to find you answers. The discriminant takes the form $$7^2-4(2)(C)$$ or $$49-8C.$$ Now, simply solve this equation such that it is equal to zero, less than zero and greater than zero, and you will have your answer.

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  • $\begingroup$ is this question want me to test when c>0, c=0 and c<0 ? $\endgroup$ – carry Mar 4 '14 at 23:49
  • $\begingroup$ No. It wants you to use the equation $b^2-4ac$ to test whether the equation $2x^2+7x+C=0$ has one, zero, or two solutions. The determinant gives you that information. $\endgroup$ – user124862 Mar 4 '14 at 23:50
  • $\begingroup$ thanks for the hint~ i don't really get the idea but now i get it $\endgroup$ – carry Mar 5 '14 at 0:02

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