Help again

Printable View

July 30th 2006, 02:38 PM
Chester

Help again

Please explain how to solve this.

The product of two consecutive even integers is 168. Find the two integers.
Hint: Assume the first integer is x
Set up the equation first that satisfies the given condition. Then solve the equation.

This is the equation that i came up with. Ican t seem to get past

x(x+2)=168

x^2+2x-168=0 (i cant get past here)

Thanks,
Chester
July 30th 2006, 02:40 PM
ThePerfectHacker

Quote:

Originally Posted by Chester

x(x+2)=168

x^2+2x-168=0 (i cant get past here)

Good job.
Find some factors of 168: 12 and 14
which diffrence is 2 thus,
$(x+14)(x-12)=168$
Thus,
$x=-14,x=12$
July 30th 2006, 02:44 PM
galactus

You could always fall back on the quadratic formula. I'd rather factor.

What 2 numbers when added equal 2 and when multiplied equal -168.

Let's see.......how about.....................-12 and 14. (-12)(14)=-168

-12+14=2

$x^{2}-12x+14x-168$

$(x^{2}-12x)+(14x-168)$

$x(x-12)+14(x-12)$

(x-12)(x+14)
July 30th 2006, 02:59 PM
Soroban

Hello, Chester!

You did an excellent on job on the hard part: setting up the equation.
So where is your difficulty?

I assume you know to solve a quadratic equation
. . and that you know how to factor.
Did you run of factorings to try?

$x^2+2x-168\:=\:0$

We want two numbers with a product of 168 and a difference of 2.

Start dividing 168 by 1, 2, 3, . . .

$168 \div 1 = 168\quad\Rightarrow\quad 1\cdot168$

$168 \div 2 = 84\quad\Rightarrow\quad 2\cdot84$

$168 \div 3\;\;\text{ not exact}$

$168 \div 4 = 42\quad\Rightarrow\quad 4\cdot42$

$168 \div 5,\; 168 \div 6,\;\168 \div 7\;\;\text{ not exact}$

$168 \div 8 = 21\quad\Rightarrow\quad 8\cdot21$

$168 \div 9,\;168 \div 9,\;168\div 10,\;168\div11\;\;\text{ not exact}$

$168 \div 12 = 14\quad\Rightarrow\quad 12\cdot14\;\;\leftarrow$ There! a difference of 2

Therefore: . $x^2 + 2x - 168 \;= \;(x - 12)(x + 14)$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Of course, you can use the Quadratic Formula for factoring . . .
July 30th 2006, 03:18 PM
Chester

Thanks

Thanks again everyone.

I really appreciate the help!

Chester
July 30th 2006, 06:10 PM
Quick

I believe the answer is $12\cdot14$ not $(-12)\cdot14$
notice that negative 12 and 14 are not consecutive even integers.
July 30th 2006, 06:18 PM
ThePerfectHacker

Quote:

Originally Posted by Quick

I believe the answer is $12\cdot14$ not $(-12)\cdot14$
notice that negative 12 and 14 are not consecutive even integers.

There two answers for $x$ .
$x=12, x+2=14$ and $12\times 14=168$
And,
$x=-14,x+2=-12$ and $-14\times -12=168$