The sum of the square of two consecutive positive odd integers is 202, how do you find the integers?

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The sum of the square of two consecutive positive odd integers is 202, how do you find the integers?

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Answer 1 · 2016-02-08T12:26:25

9 , 11

Explanation:

let n be a positive odd integer

then the next consecutive odd number is , n + 2 , since odd numbers have a difference of 2 between them.

from the given statement : $n^{2} + {(n + 2)}^{2} = 202$ $n^2 + (n+2)^2 = 202$

expanding gives : $n^{2} + n^{2} + 4 n + 4 = 202$ $n^2 + n^2 + 4n + 4 =202$

this is a quadratic equation so collect terms and equate to zero.

$2 n^{2} + 4 n - 198 = 0$ $2n^2 + 4n -198 = 0$

common factor of 2 : $2 (n^{2} + 2 n - 99) = 0$ $2(n^2 + 2n - 99) = 0$

now consider factors of -99 which sum to +2. These are 11 and -9.

hence : 2(n + 11 )(n-9 ) = 0

(n + 11 ) = 0 or (n-9) = 0 which leads to n = -11 or n = 9

but n > 0 hence n = 9 and n+ 2 = 11

Answer 2 · 2016-02-08T12:41:43

Always remember that $o d d$ $color(blue)(odd$ $n$ $color(blue)(n$ $u m b e r s$ $color(blue)(umbers$ always differ in the value of $2$ $color(green)2$

So,let the first number be $x$ $color(red)(x$

Then the second number will be $x + 2$ $color(red)(x+2$

Then,

${(x)}^{2} + {(x + 2)}^{2} = 202$ $color(green)((x)^2+(x+2)^2=202$

Use formula ${(a + b)}^{2} = a^{2} + 2 a b + b^{2}$ $color(green)((a+b)^2color(blue)(=color(brown)(a^2+2ab+b^2$

$\to x^{2} + x^{2} + 2 x (2) + 2^{2} = 202$ $rarrcolor(green)(x^2+x^2+2x(2)+2^2=color(blue)(202$

$\to x^{2} + x^{2} + 4 x + 4 = 202$ $rarrcolor(green)(x^2+x^2+4x+4=color(blue)(202$

$\to 2 x^{2} + 4 x + 4 = 202$ $rarrcolor(green)( 2x^2+4x+4=color(blue)(202$

$\to 2 x^{2} + 4 x = 202 - 4$ $rarrcolor(green)( 2x^2+4x=color(blue)(202-4$

$\to 2 x^{2} + 4 x = 198$ $rarr color(green)(2x^2+4x=color(blue)(198$

$\to 2 x^{2} + 4 x - 198 = 0$ $rarrcolor(green)( 2x^2+4x-198=color(blue)(0$

Now this is a Quadratic equation (in form $a x^{2} + b x^{2} + c = 0$ $color(brown)(ax^2+bx^2+c=0$ ) So,we can use the Quadratic formula or factor it out.

Luckily,we can factor it to

$\to 2 x^{2} + 4 x - 198 = (2 x + 22) (a - 9) = 0$ $rarrcolor(green)( 2x^2+4x-198=color(brown)((2x+22)(a-9)=0$

$\to (2 x + 22) (a - 9) = 0$ $rarrcolor(green)( (2x+22)(a-9)=color(brown)(0$

Now we have two values for $x$ $color(green)(x$ which are

$1) \to x = - \frac{22}{2} = - 11$ $1)rarr color(green)(x=-22/2=-11$

$2) \to x = 9$ $2)rarrcolor(green)( x=9$

Now we need to find $x + 2$ $color(orange)(x+2$

If $x = - 11$ $color(green)(x=-11$

Then, $x + 2 = - 11 + 2 = - 9$ $color(orange)(x+2=-11+2=-9$

And if $x = 9$ $color(green)(x=9$

Then, $x + 2 = 9 + 2 = 11$ $color(orange)(x+2=9+2=11$

So,at the end we conclude the if the first integer is $- 11$ $color(green)(-11$ then second integer is $- 9$ $color(orange)(-9$ and if the first integer is $9$ $color(green)(9$ ,the second integer is $11$ $color(orange)(11$ .

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The sum of the square of two consecutive positive odd integers is 202, how do you find the integers?

2 Answers

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