If p and q are nonzero integers, which pair of points must lie in the same quadrant?

1 answer:

5 0

I have the feeling that there are some choices that are supposed to
go with this question, but apparently you decided not to share them
for some reason.

'p' and 'q' are non-zero integers.
In order to show you groups of points that are all in the same quadrant,
I need another pair of non-zero numbers. Call them 'A' and 'B'. They
don't need to be integers; they just need to be positive.

-- Any points of the form (Ap, Bq) are all in the First quadrant.

-- Any points of the form (-Ap, Bq) are all in the Second quadrant.

-- Any points of the form (-Ap, -Bq) are all in the Third quadrant.

-- Any points of the form (Ap, -Bq) are all in the Fourth quadrant.

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Alecsey [184]

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8 0

3 years ago

How can i solve by completing the square 4r^2-28r=-49

Natalija [7]

To complete the square, the second degree term must have a coefficient of 1.
Since the second degree term here has a coefficient of 4, we start by dividing each term on both sides by 4.

$4r^2 - 28r = -49$

$\dfrac{4r^2}{4} - \dfrac{28}{4}r = -\dfrac{49}{4}$

$r^2 - 7r = -\dfrac{49}{4}$

Now we can complete the square.
First, we need to find what number completes the square.
We take the coefficient of the first degree term, -7 in this case.
Divide it by 2 and square it. -7 divided by 2 is the fraction -7/2.
Now we square -7/2 to get 49/4.
We add 49/4 to both sides.

$r^2 - 7r + \dfrac{49}{4} = -\dfrac{49}{4} + \dfrac{49}{4}$

$(r - \dfrac{7}{2})^2 = 0$

$r - \dfrac{7}{2} = 0$

$r = \dfrac{7}{2}$