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inysia [295]
3 years ago
11

G 1) a2 + 3a – 40 = 0 . solving quadratic equations

Mathematics
1 answer:
QveST [7]3 years ago
5 0

Answer:

The solution set is {-8, 5}.

Step-by-step explanation:

a^2 + 3a – 40 = 0 .

We need 2 numbers whose product is  -40 and whose sum is +3.

They would be + 8 and - 5, so we have the factors:

(a + 8)(a - 5) = 0

Now either a + 8 = 0 or a - 5 = 0, so:

a = -8 or 5.

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\bf \textit{hyperbolas, horizontal traverse axis } \\\\ \cfrac{(x- h)^2}{ a^2}-\cfrac{(y- k)^2}{ b^2}=1 \qquad  \begin{cases} center\ ( h, k)\\ vertices\ ( h\pm a,  k)\\ c=\textit{distance from}\\ \qquad \textit{center to foci}\\ \qquad \sqrt{ a ^2 + b ^2} \end{cases}\\\\ -------------------------------


\bf \cfrac{(x-5)^2}{144}-\cfrac{(y-4)^2}{81}=1\implies \cfrac{(x-5)^2}{12^2}-\cfrac{(y-4)^2}{9^2}=1 \\\\\\ \begin{cases} h=5\\ k=4\\ a=12\\ b=9 \end{cases}\implies c=\sqrt{144+81}\implies c=\sqrt{225}\implies c=15 \\\\\\ \stackrel{center}{(5,4)}\qquad \qquad \stackrel{\textit{because is a horizontal traverse hyperbola}}{\stackrel{foci}{\stackrel{(5\pm 15,4)}{(20,4),(-10,4)}}\qquad \qquad \stackrel{vertices}{\stackrel{(5\pm 12,4)}{(17,4),(-7,4)}}}

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