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Find the product of all positive integer values of $c$ such that $3x^2+7x+c=0$ has two real roots. I will award a lot of points

Respuesta :

tramserran

Answer:  24

Step-by-step explanation:

Let's find one solution:

3x² + 7x + c = 0

a=3 b=7  c=c

First, let's find c so that it has REAL ROOTS.

⇒ Discriminant (b² - 4ac) ≥ 0

                         7² - 4(3)c ≥ 0

                         49 - 12c ≥ 0

                               -12c  ≥ -49

                                [tex]c\leq\dfrac{-49}{-12}\quad \rightarrow c\leq \dfrac{49}{12}[/tex]      

Since c must be a positive integer, 1 ≤ c ≤ 4

Example: c = 4

3x² + 7x + 4 = 0

(3x + 4)(x + 1) = 0

x = -4/3, x = -1         Real Roots!

You need to use Quadratic Formula to solve for c = {1, 2, 3}

Valid solutions for c are: {1, 2, 3, 4)

Their product is: 1 x 2 x 3 x 4 = 24

Answer:

$3x^2+7x+c=0$

comparing above equation with ax²+bx+c

a=3

b=7

c=1

using quadratic equation formula

[tex]x = \frac{ - b + - \sqrt{b {}^{2} - 4ac} }{ 2a} [/tex]

x=(-7+-√(7²-4×3×1))/(2×3)

x=(-7+-√13)/6

taking positive

x=(-7+√13)/6=

taking negative

x=(-7-√13)/6=

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