The product of two positive consecutive even integers is 224. How do you find the integers?

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The product of two positive consecutive even integers is 224. How do you find the integers?

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Answer 1 · 2016-10-20T20:33:26

The two consecutive positive integers whose product is $224$ $224$ are $14 and 16$ $color(blue)(14 and 16)$

Explanation:

Let the first integer be $x$ $color(blue)x$
since the second is the consecutive even then, it is $x + 2$ $color(blue)(x+2)$

The product of these integers is $224$ $224$ i.e if we multiply $x$ $color(blue)x$ and $x + 2$ $color(blue)(x+2)$ the result is $224$ $224$ that is:

$x \cdot x + 2 = 224$ $color(blue)x*color(blue)(x+2)=224$
$\Rightarrow x^{2} + 2 x = 224$ $rArrx^2+2x=224$
$\Rightarrow x^{2} + 2 x - 224 = 0$ $rArrcolor(green)(x^2+2x-224=0)$

Let us compute the quadratic roots:

$δ = b^{2} - 4 a c = 4^{2} - 4 (1) (- 224) = 4 + 896 = 900$ $color(brown)(delta=b^2-4ac)=4^2-4(1)(-224)=4+896=900$
$x_{1} = \frac{- b - \sqrt{δ}}{2 a} = \frac{- 2 - \sqrt{900}}{2 \cdot 1} = \frac{- 2 - 30}{2} = (- \frac{32}{2}) = - 16$ $color(brown)(x_1=(-b-sqrtdelta)/(2a))=(-2-sqrt900)/(2*1)=(-2-30)/2 =(-32/2)=-16$
$x_{2} = \frac{- b + \sqrt{δ}}{2 a} = \frac{- 2 + \sqrt{900}}{2 \cdot 1} = \frac{- 2 + 30}{2} = (\frac{28}{2}) = 14$ $color(brown)(x_2=(-b+sqrtdelta)/(2a))=(-2+sqrt900)/(2*1)=(-2+30)/2 =(28/2)=14$

$\Rightarrow x^{2} + 2 x - 224 = 0$ $rArrcolor(green)(x^2+2x-224=0)$
$\Rightarrow (x + 16) (x - 14) = 0$ $rArr(x+16)(x-14)=0$

Therefore,
(hint : $g i v e n x > 0$ $color(red) (given x>0)$ )

$x + 16 = 0 \Rightarrow x = - 16 r e j e c t e d$ $x+16=0rArrx=-16color(red)(rejected)$
Or
$x - 14 = 0 \Rightarrow x = 14$ $x-14=0rArrx=14$ ACCEPTED

Therefore,
The first positive integer is:
$x = 14$ $color(blue)(x=14)$
The first positive integer is:
$x + 2 = 16$ $color(blue)(x+2=16)$

The two consecutive positive integers whose product is $224$ $224$ are $14 and 16$ $color(blue)(14 and 16)$

Answer 2 · 2016-10-20T20:38:55

$14 \times 16 = 224$ $14xx16 = 224$

Explanation:

Integral to solving questions like this is an understanding of the factors of a number and what they tell us.

Consider the factors of 36:

$F_{36} = 1, 2, 3, 4, 6, 9, 12, 18, 36$ $F_36 = 1 , 2 , 3 , 4 , 6 , 9 , 12 , 18 , 36$
$\times \times \times \times \times ↑$ $color(white)(xxxxxxxxxx)uarr$

Note the following:

There are factor pairs. Each small factor is paired with a big factor.
As one increases, the other decreases.
The difference between the factors decreases as we work inwards

$1 \times 36$ $1xx36" "$ difference is 35
$2 \times 18$ $2xx18" "$ difference is 16
$3 \times 12$ $3xx12" "$ difference is 9
$4 \times 9$ $4xx9" "$ difference is 5
$6$ $6" "$ difference is 0

However, there is only ONE factor in the middle. This is because 36 is a square and the middle factor is its square root.
$\sqrt{36} = 6$ $sqrt36 = 6$
The smaller the difference between the factors of any number, the closer they are to the square root.

Now for this question ..... The fact that the even numbers are consecutive means that they are very close to the square root of their product.

$sqrt224 = 14.966629.....$

Try the even numbers closest to this number. One a bit more, the other a bit less. We find that ...............

$14xx16 = 224$

These are the numbers we are looking for.
They lie on either side of $sqrt224$

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The product of two positive consecutive even integers is 224. How do you find the integers?

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