+0

# Could I please have some help these two problems? They're due on Wed, Sept. 4

0
137
1

1. There are 5 quadratics below. Four of them have two distinct roots each. The other has only one distinct root; find the value of that root.

4x^2+16x-9

2x^2+80x+400

x^2-6x-9

4x^2-12x+9

-x^2+14x+49

2. How many ordered pairs of positive integers satisfy the equation x/y=225/xy+y/x?

Sep 1, 2019

#1
+2

So I'm not completely doing your homework for you, I'll just get you started.

1. To find out what type of roots a quadratic in the form ax2+bx+c, plug the coeficients into the discriminant: b2-4ac

If this is positive, there are two real solitutions. If it is negative, there are two non real solitutions. If it is 0, then there is one real solitution. You want to find the discriminants of the 5 quadratics to find the discriminant that is 0.

I'll do the first one:

a=4, b=16 and c=-9

162-4(4)(-9)=256-(-144)=400, so the first one is not the answer.

2. I'm going to interperate this one how I think you meant to write it (you didn't put parenthisies around the xy so the y should be in the numerator): $$\frac{x}{y}=\frac{225}{xy}+\frac{y}{x}$$ (and this interperation works out nicely)

First I would start by multiplying by xy to remove the varaibles from the denominator:

x2=225+y2

x2-y2=225

(x+y)(x-y)=225

x and y will be integers when (x+y) and (x-y) are factor pairs of 225

The factors of 225 are:

1 and 225

3 and 75

5 and 45

ect.

For every factor pair there will be integers x and y that satisfy the equation

List the remaining factor pairs until you get to 225 and 1 and count them to get the answer.

To find the solitutions that go with the factors 1 and 225 (which you don't need to do), you would solve x+y=1 and x-y=225

Sep 1, 2019

#1
+2

So I'm not completely doing your homework for you, I'll just get you started.

1. To find out what type of roots a quadratic in the form ax2+bx+c, plug the coeficients into the discriminant: b2-4ac

If this is positive, there are two real solitutions. If it is negative, there are two non real solitutions. If it is 0, then there is one real solitution. You want to find the discriminants of the 5 quadratics to find the discriminant that is 0.

I'll do the first one:

a=4, b=16 and c=-9

162-4(4)(-9)=256-(-144)=400, so the first one is not the answer.

2. I'm going to interperate this one how I think you meant to write it (you didn't put parenthisies around the xy so the y should be in the numerator): $$\frac{x}{y}=\frac{225}{xy}+\frac{y}{x}$$ (and this interperation works out nicely)

First I would start by multiplying by xy to remove the varaibles from the denominator:

x2=225+y2

x2-y2=225

(x+y)(x-y)=225

x and y will be integers when (x+y) and (x-y) are factor pairs of 225

The factors of 225 are:

1 and 225

3 and 75

5 and 45

ect.

For every factor pair there will be integers x and y that satisfy the equation

List the remaining factor pairs until you get to 225 and 1 and count them to get the answer.

To find the solitutions that go with the factors 1 and 225 (which you don't need to do), you would solve x+y=1 and x-y=225

power27 Sep 1, 2019