Question #313e5

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Question #313e5

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Answer 1 · 2016-08-03T15:07:44

The reqd. larger even $+ v e$ $+ve$ integer is $10$ $10$ .

Explanation:

Let the reqd. first even integer be $2 x$ $2x$ , then, observe that, the next to this even integer will be $2 x + 2$ $2x+2$ and not $2 x + 1$ $2x+1$ .

Now, "The square of the sum of 2 consecutive positive even integers" $= {[(2 x) + (2 x + 2)]}^{2}$ $=[(2x)+(2x+2)]^2$ . Let us call this $A$ $A$ .

Next, "the sum of their squares" $= [{(2 x)}^{2} + {(2 x + 2)}^{2}]$ $=[(2x)^2+(2x+2)^2]$ . Call it $B$ $B$ .

We are given that, $A > B$ $A > B$ by $160$ $160$ , i.e., $A = B + 160$ $A=B+160$ .

Hence, we have,

${[(2 x) + (2 x + 2)]}^{2} = [{(2 x)}^{2} + {(2 x + 2)}^{2}] + 160$ $[(2x)+(2x+2)]^2=[(2x)^2+(2x+2)^2]+160$

${(4 x + 2)}^{2} = 4 x^{2} + 4 {(x + 1)}^{2} + 160$ $(4x+2)^2=4x^2+4(x+1)^2+160$

$4 {(2 x + 1)}^{2} = 4 x^{2} + 4 {(x + 1)}^{2} + 160$ $4(2x+1)^2=4x^2+4(x+1)^2+160$ .

Dividing throughout by $4$ $4$ and then simplifying, we get,

$4 x^{2} + 4 x + 1 = x^{2} + x^{2} + 2 x + 1 + 40$ $4x^2+4x+1=x^2+x^2+2x+1+40$ , or,

$2 x^{2} + 2 x - 40 = 0 \Rightarrow x^{2} + x - 20 = 0$ $2x^2+2x-40=0 rArr x^2+x-20=0$

$\Rightarrow (x + 5) (x - 4) = 0 \Rightarrow x = - 5, or, x = 4$ $rArr (x+5)(x-4)=0 rArr x=-5, or, x=4$ .

since we need $+ v e$ $+ve$ integers, $x \neq - 5$ $x!=-5$ .

$x = 4 \Rightarrow 2 x = 8, 2 x + 2 = 10$ $x=4 rArr 2x=8, 2x+2=10$

Hence, the reqd. larger even $+ v e$ $+ve$ integer is $10$ $10$ .

Hope, this'll help! Enjoy Maths.!

Answer 2 · 2016-08-03T16:44:59

The other solutions are correct. This is a very slightly different approach. Upon reflection, it is virtually the same as that by Ratnaker

The larger number is 10

Explanation:

The two numbers are even so my starting point is to make sure the first integer is even (divisible by 2)

Let the 'seed value' (can be even or odd) be $n$ $n$

Then $2 n$ $2n$ is even

Let the first number be $2 n$ $2n$
Let the second number be $2 n + 2$ $2n+2$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$Breaking the question down into its component parts$ $color(blue)("Breaking the question down into its component parts")$

The square of the sum of 2 consec. numb. $\to {[2 n + (2 n + 2)]}^{2}$ $->[2n+(2n+2)]^2$

is greater than $\to {[2 n + (2 n + 2)]}^{2} > ?$ $->[2n+(2n+2)]^2>?$

the sum of their squares $\to {[2 n + (2 n + 2)]}^{2} > {(2 n)}^{2} + {(2 n + 2)}^{2}$ $->[2n+(2n+2)]^2>(2n)^2+(2n+2)^2$

by 160 $\to {[2 n + (2 n + 2)]}^{2} = {(2 n)}^{2} + {(2 n + 2)}^{2} + 160$ $color(blue)(->[2n+(2n+2)]^2=(2n)^2+(2n+2)^2+160)$
$↑$ $" "color(brown)(uarr)$
$Changes the > to = by fixing their difference$ $color(brown)("Changes the > to = by fixing their difference")$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$\Rightarrow {(4 n + 2)}^{2} = 4 n^{2} + {(2 n + 2)}^{2} + 160$ $=>(4n+2)^2=4n^2+(2n+2)^2+160$
$=>16n^2+16n+cancel(4)=4n^2+4n^2+8n+cancel(4)+160$

$=>16n^2+16n=8n^2+8n+160$

$=>8n^2+8n-160=0$

Divide by 8
$=>n^2+n-20=0$

$(n-4)(n+5)=0$

$n=+4 or n=-5$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$n=-5$ is discounted as our numbers are positive

$color(brown)("Remember that "n" is a 'seed value'")$

$=> "first number is "2n=2(4)=8$

$color(blue)("So the larger number is "8+2=10)$

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Question #313e5

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