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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

recursive recursion
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carry
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4 Answers4

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Hint: use the discriminant to find the number of roots.

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

Newb
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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.

Stella Biderman
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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.

  • is this question want me to test when c>0, c=0 and c<0 ? – carry Mar 04 '14 at 23:49
  • 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. –  Mar 04 '14 at 23:50
  • thanks for the hint~ i don't really get the idea but now i get it – carry Mar 05 '14 at 00:02
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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.
Jimmy R.
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