Determine the number of roots, using the most efficient method possible.

Mathematics

1 answer:

Dmitry [639]3 years ago

7 0

Answer:

2 roots in every case, not all are real.

Step-by-step explanation:

All of these equations are quadratic equations (degree 2). Every quadratic has two roots. They may be identical (looks like 1 root), and they may be complex (zero real roots), but there are always 2 of them.

a) the y-value of the vertex is negative and the parabola opens downward (leading coefficient -5), so there are no real zeros and both roots are complex.

b) each binomial factor contributes a root. Both roots are real.

c) the discriminant is positive, (3²-4·2·1=1), so both roots are real.

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nadya68 [22]

The answer that would be correct is C.

4 0

3 years ago

-3,6,-9,12,-15 which of the following represents the general term for the sequence give?

Mademuasel [1]

The general term for the sequence $a_{n}=(-1)^{n} 3 n$

<h3><u>Solution:</u></h3>

Given sequence is -3, 6, -9, 12, -15

We have to find the general term of sequence

The terms in the sequence are found out by using the recursive definition:

$a_{n}=(-1)^{n} 3 n$

Let us use this definition to find out the terms and check if it matches our given sequence

$\begin{array}{l}{\text { For } n=1:-(-1)^{n} 3 n=(-1)^{1} \times 3 \times 1=-3} \\\\ {\text { For } n=2:-(-1)^{n} 3 n=(-1)^{2} \times 3 \times 2=1 \times 6=6} \\\\ {\text { For } n=3:-(-1)^{n} 3 n=(-1)^{3} \times 3 \times 3=-1 \times 9=-9} \\\\ {\text { For } n=4:-(-1)^{n} 3 n=(-1)^{4} \times 3 \times 4=1 \times 12=12} \\\\ {\text { For } n=5:-(-1)^{n} 3 n=(-1)^{5} \times 3 \times 5=-1 \times 15=-15}\end{array}$