I have 5 questions...

1) The sum of two numbers is 12. Their product is 36. Find the two numbers.

2) The sum of two numbers is -15. Their product is -100. Find the two numbers.

3) The sum of two numbers is -8. Their product is 12. Find the two numbers.

4) The length of a rectangle is four times the width. The perimeter is 150 inches. Find the dimensions of the rectangle.

5) The length of a rectangle is four less than three times the width. The area is 20 square cm. Find the dimensions of the rectangle.

We will be happy to critique your thinking.

q1.6,6

25

Sure! Let's break down each question and figure out the answers step-by-step.

1) The sum of two numbers is 12. Their product is 36. Find the two numbers.

To solve this problem, we need to set up a system of equations. Let's call the two numbers x and y. According to the problem, their sum is 12:

x + y = 12

Their product is given as 36:

xy = 36

We can now solve this system of equations using various methods such as substitution or elimination. Let's use substitution here. Solving the first equation for x, we get:

x = 12 - y

Now we substitute this value of x into the second equation:

(12 - y)y = 36

Simplifying further, we have:

12y - y^2 = 36

Rearranging and putting the equation in standard form:

y^2 - 12y + 36 = 0

This quadratic equation factors as:

(y - 6)(y - 6) = 0

So, y = 6.

Substituting this value back into the first equation:

x + 6 = 12
x = 6

Therefore, the two numbers are 6 and 6.

2) The sum of two numbers is -15. Their product is -100. Find the two numbers.

Following the same steps as before, let's call the two numbers x and y. According to the problem, their sum is -15:

x + y = -15

Their product is given as -100:

xy = -100

Using substitution again, we solve the first equation for x:

x = -15 - y

Substituting this value of x into the second equation:

(-15 - y)y = -100

Distributing the -1:

-15y - y^2 = -100

Arranging in standard form:

y^2 + 15y - 100 = 0

Factoring this quadratic equation may not be immediately obvious, so we can use the quadratic formula:

y = (-15 ± sqrt(15^2 - 4*1*(-100))) / (2*1)

Simplifying further:

y = (-15 ± sqrt(225 + 400)) / 2
y = (-15 ± sqrt(625)) / 2
y = (-15 ± 25) / 2

We get two possible values for y:

y1 = (-15 + 25) / 2 = 5
y2 = (-15 - 25) / 2 = -20

Substituting these values back into the first equation, we find the corresponding values of x:

For y = 5:
x + 5 = -15
x = -20

For y = -20:
x + (-20) = -15
x = 5

Therefore, the two possible pairs of numbers are (-20, 5) and (5, -20).

3) The sum of two numbers is -8. Their product is 12. Find the two numbers.

Following the same process, we'll call the two numbers x and y. According to the problem, their sum is -8:

x + y = -8

Their product is given as 12:

xy = 12

Again, solving the first equation for x:

x = -8 - y

Substituting this value of x into the second equation:

(-8 - y)y = 12

Distributing the -1 and arranging in standard form:

y^2 + 8y + 12 = 0

This quadratic equation factors as:

(y + 2)(y + 6) = 0

So, y can be either -2 or -6.

Substituting these values back into the first equation, we find the corresponding values of x:

For y = -2:
x + (-2) = -8
x = -6

For y = -6:
x + (-6) = -8
x = -2

Therefore, the two possible pairs of numbers are (-6, -2) and (-2, -6).

4) The length of a rectangle is four times the width. The perimeter is 150 inches. Find the dimensions of the rectangle.

Let's call the width of the rectangle x. According to the problem, the length is four times the width:

Length = 4x

The perimeter of a rectangle is the sum of all the sides, so we can set up an equation:

Perimeter = 2(Length + Width)

Substituting the values, we have:

150 = 2(4x + x)

Simplifying further:

150 = 2(5x)
150 = 10x
x = 15

Therefore, the width of the rectangle is 15 inches, and the length is four times that, which is 60 inches.

5) The length of a rectangle is four less than three times the width. The area is 20 square cm. Find the dimensions of the rectangle.

Similarly, let's call the width of the rectangle x. According to the problem, the length is four less than three times the width:

Length = 3x - 4

The area of a rectangle is the product of the length and width, so we can set up an equation:

Area = Length * Width

Substituting the values, we have:

20 = (3x - 4)x

Simplifying further:

20 = 3x^2 - 4x

Rearranging and putting the equation in standard form:

3x^2 - 4x - 20 = 0

As this quadratic equation doesn't seem to factor easily, we can use the quadratic formula:

x = (-(-4) ± sqrt((-4)^2 - 4*3*(-20))) / (2*3)

Simplifying further:

x = (4 ± sqrt(16 + 240)) / 6
x = (4 ± sqrt(256)) / 6
x = (4 ± 16) / 6

We get two possible values for x:

x1 = (4 + 16) / 6 = 20/6 = 10/3
x2 = (4 - 16) / 6 = -12/6 = -2

Since the dimensions of a rectangle cannot be negative, we discard x2. Therefore, the width of the rectangle is 10/3 cm, and the length can be calculated using the given formula:

Length = 3(10/3) - 4 = 10 - 4 = 6 cm

Therefore, the width of the rectangle is 10/3 cm, and the length is 6 cm.

I hope this helps! Let me know if you have any further questions or if there's anything else I can assist you with.