Question

An integer is 2 less than 4 times another. If the product of the two integers is 56, then find the integers

Answer 1

-7/2 and 4

Explanation:

start by using x to represent the number

the first number is x and the 2nd number is 4 times x minus 2

these numbers can be represented as x & 4x-2

now since the product of the 2 numbers is 56, set x times 4x-2 equal to 56

when you expand the equation you get

4x^2-2x=56

subtract 56 from the right and you have a quadratic equation. simply factor and you get the roots -7/2 and 4

Answer 2

x = 4

y = 14

Explanation:

Step one : we interpret the word problem using equations

Let's assume the integer is y;

and another integer is x

4 times x = 4x

2 less than 4 times x = 4x - 2

The first integer, y = 4x - 2

The product of x and y is 56

y = 4x - 2

x(4x - 2) = 56

On expansion,

4x² - 2x = 56

The general form of a quadratic equation is

ax² + bx + c = 0

Step 2 : represent the equation, 4x² - 2x = 0 in the form of a quadratic equation by subtracting 56 from both sides

4x² - 2x - 56 = 56 - 56

4x² - 2x - 56 = 0

Now we have a quadratic equation

Step 3 : solve the quadratic equation, 4x² - 2x - 56 = 0

ax² + bx + c = 0

a x c = ac

We find factors of ac that can be added together to give b

4 x -56 = -224

Factors of -224 are

1 X -224 = -224

2 X -112 = -224

.

.

.

-16 X 14 = -224

-16 + 14 = -2

-16x + 14x = -2x

4x² - 16x + 14x - 56 = 0

(4x² - 16x) + (14x - 56) = 0

Factorise each bracket

4x(x - 4) + 14 ( x - 4) = 0

(4x + 14)(x - 4) = 0

4x + 14 = 0

x - 4 = 0

x = -14/4

Or

x = 4

Since the numbers are integers and not fractions, x = 4 is the valid solution for the statement.

The integer, y that is 2 less than 4 times another integer x = (4 times 4) - 2 = 16 - 2 = 14