How do you find two consecutive odd integers such that the square of the first, added to 3 times the second is 24?

Question

How do you find two consecutive odd integers such that the square of the first, added to 3 times the second is 24?

1 Answer

Impact of this question

3750 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-09-29T17:18:37

I found: $3 and 5$ $3 and 5$

Explanation:

Let us call our odd integers:
$2 n + 1$ $2n+1$ and $2 n + 3$ $2n+3$ so we have:
${(2 n + 1)}^{2} + 3 (2 n + 3) = 24$ $(2n+1)^2+3(2n+3)=24$
$4 n^{2} + 4 n + 1 + 6 n + 9 - 24 = 0$ $4n^2+4n+1+6n+9-24=0$
$4 n^{2} + 10 n - 14 = 0$ $4n^2+10n-14=0$
Using the Quadratic Formula to solve for $n$ $n$ we have:
$n_{1} = 1$ $n_1=1$ so that the two integers will be:
$2 + 1 = 3$ $2+1=3$ and $2 + 3 = 5$ $2+3=5$
or:
$n_{2} = - \frac{7}{2}$ $n_2=-7/2$ so that the two integers will be:
$2 (- \frac{7}{2}) + 1 = - 6$ $2(-7/2)+1=-6$ and $2 (- \frac{7}{2}) + 3 = - 4$ $2(-7/2)+3=-4$ but these are even!

Science

Math

Humanities

... and beyond

How do you find two consecutive odd integers such that the square of the first, added to 3 times the second is 24?

1 Answer

Explanation:

Related questions

Impact of this question