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Find the rational roots of

Find the rational roots of-example-1

2 Answers

2 votes

Answer:

C

Explanation:

Remark

You need to test the discriminate to see what happens. The discriminate must be a perfect square. It must also be minus. As a random choice I'm going to do B very carefully and then C

Choice B

Givens

2x^2 + x + 17

a = 2

b = 1

c = 17

Formula

b^2 - 4*a*c

Solution

(1^2 - 4*2*17)

1 - 136

- 135

Now is - 135 a perfect square? Put it into it's prime factors to find out.

-135: 3*3*3*5 It is not a perfect square. It has an odd number of 3s (There are 3 of them) and an odd number of 5s (0nly 1).

sqrt(-135) = sqrt(-3*3*3*5) You can take two threes out. 3*sqrt(-3*5)= 3*sqrt(-15) which is not a perfect square. You need go no further. B is not the answer.

Choice C

Givens

x^2 + 2x + 17

a = 1

b = 2

c = 17

Formula

b^2 - 4*a*c

Solution

(2^2 - 4*1*17)

4 - 68

- 64 Now do you see that 64 is a perfect square? Except for the minus sign.

sqrt (-64) = 8i

So now you you use the quadratic formula substituting 8i for the discriminate.

x = [- b +/- discriminate ] / 2a

x1 =[-2 + discriminate] / 2(1)

x1 = -2/2 + discriminate/2

x1 = -1 + 8i/2

x1 = -1 + 4i

x2 = - 1 - 4i using the same steps as above. The answer is C

Comments about the other three choices.

A: gives a perfect square for the discriminate but it's too small. It is 2i

B: has been solved

C: is the answer

D: gives sqrt(-15) which is not a perfect square. As an exercise you should try D


User Naree
by
8.6k points
1 vote

Answer:

x² +2x +17 = 0

Explanation:

When a root is "a", the corresponding factor of the polynomial is (x - a). Here, that means the polynomial equation will be ...

... (x -(-1 +4i))·(x -(-1 -4i)) = 0

... ((x+1) -4i)·((x+1) +4i) = 0

These factors correspond to the factors of the difference of two squares, so the equation can be written as ...

... (x+1)² - (4i)² = 0 . . . . . . written as the difference of two squares

... x² +2x +1 -(-16) = 0 . . . each of the squares evaluated

... x² +2x +17 = 0 . . . . . . . simplified form

User Danny Tuppeny
by
7.8k points

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