Question

1. Two numbers, which differ by 3, have a product of 88.
Find them.​

Answer 1

First Set: [11, 8], Second Set: [-8, -11]

Explanation:

The first number will be x. The second number is x-3. So now make an equation: x(x-3)=88 because their product equals 88. So now solve.

Solving:

`(x)(x-3)=88`

`x^(2)-3x-88=0`

Now you can factor:

`(x-11)(x+8)=0`

`x=11, x=-8`

Now plug in each number for x for the following equation:

`(x)(x-3)=88`

`(11)(11-3)=88`

`88=88`

This is true.

`(-8)(-8-3)=88`

`88=88`

This is true.

Both are true. So now you have two sets of answers.

Your welcome!

Answer 2

Let x be one of the numbers

then since they differ by 3... Let (x-3) be the 2nd Number

Their product is 88

(x-3)x = 88

x²-3x =88

x² - 3x - 88=0

Solving The quadratic eqn

x=11 or x = -8.

To check this answer

since the Numbers are x and x-3

x= 11

x-3 = 11 - 3 = 8

(x-3)x = 8x11 = 88.

Or When x=-8

(x-3) = -8 - 3 = -11

x= -8

(x-3)x = -8 x - 11 = 88.

Have a great day!