How do you find the two consecutive even integers whose product is 840?

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How do you find the two consecutive even integers whose product is 840?

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Answer 1 · 2016-07-05T18:17:05

Translate the problem to an algebraic statement and solve a quadratic equation to find that there are two pairs of numbers that satisfy the problem.

Explanation:

When we are solving algebraic problems, the first thing we must do is define a variable for our unknowns. Our unknowns in this problem are two consecutive even numbers whose product is $840$ $840$ . We'll call the first number $n$ $n$ , and if they're consecutive even numbers, the next one will be $n + 2$ $n+2$ . (For example, $4$ $4$ and $6$ $6$ are consecutive even numbers and $6$ $6$ is two more than $4$ $4$ ).

We are told that the product of these numbers is $840$ $840$ . That means these numbers, when multiplied together, produce $840$ $840$ . In algebraic terms:
$n \cdot (n + 2) = 840$ $n*(n+2)=840$

Distributing the $n$ $n$ , we have:
$n^{2} + 2 n = 840$ $n^2+2n=840$

Subtracting $840$ $840$ from both sides gives us:
$n^{2} + 2 n - 840 = 0$ $n^2+2n-840=0$

Now we have a quadratic equation. We can try to factor it, by finding two numbers that multiply to $- 840$ $-840$ and add to $2$ $2$ . It might take a while, but eventually you'll find these numbers are $- 28$ $-28$ and $30$ $30$ . Our equation factors into:
$(n - 28) (n + 30) = 0$ $(n-28)(n+30)=0$

Our solutions are:
$n - 28 = 0 \to n = 28$ $n-28=0->n=28$
$n + 30 = 0 \to n = - 30$ $n+30=0->n=-30$

Thus, we have two combinations:

$28$ $28$ and $28 + 2$ $28+2$ , or $30$ $30$ . You can see that $28 \cdot 30 = 840$ $28*30=840$ .
$- 30$ $-30$ and $- 30 + 2$ $-30+2$ , or $- 28$ $-28$ . Again, $- 30 \cdot - 28 = 840$ $-30*-28=840$ .

Answer 2 · 2016-07-05T18:37:44

The reqd. nos. are $- 30, - 28$ $-30,-28$ or, $28, 30 .$ $28, 30.$

Explanation:

Suppose that the reqd. integers are $2 x$ $2x$ and $2 x + 2$ $2x+2$

By given, then, we have $2 x \cdot (2 x + 2) = 840 \Rightarrow 4 x (x + 1) = 840$ $2x*(2x+2)=840 rArr 4x(x+1)=840$ .

$:. x^2+x=840/4=210,$ or, $x^2+x-210=0$

$:. x^2+15x-14x-210=0$

$:. x(x+15)-14(x+15)=0$

$:. (x+15)(x-14)=0$
$:. x=-15, or, x=14$

CASE I

$x=-15$ , the reqd. nos. are $2x=-30, 2x+2=-28.$

Case II

$x=14$ , the. nos. are $2x=28, 2x+2=30$

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How do you find the two consecutive even integers whose product is 840?

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