Question #852f1

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Question #852f1

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Answer 1 · 2017-08-12T09:09:09

$13$ $13$ $&$ $&$ $14$ $14$

Explanation:

Let the two consecutive positive integers be $x$ $x$ $&$ $&$ $x + 1 .$ $x+1.$

$x^{2} + {(x + 1)}^{2} = 365$ $x^2+(x+1)^2=365$

We know that ${(a + b)}^{2} = a^{2} + 2 a b + b^{2} .$ $(a+b)^2=a^2+2ab+b^2.$ So the equation becomes

$\Rightarrow x^{2} + (x^{2} + 2 \cdot x \cdot 1 + 1^{2}) = 365$ $=>x^2+(x^2+2*x*1+1^2)=365$

$\Rightarrow x^{2} + x^{2} + 2 \cdot x \cdot 1 + 1^{2} = 365$ $=>x^2+x^2+2*x*1+1^2=365$

$\Rightarrow x^{2} + x^{2} + 2 x + 1 = 365$ $=>x^2+x^2+2x+1=365$

$\Rightarrow 2 x^{2} + 2 x + 1 = 365$ $=>2x^2+2x+1=365$

$\Rightarrow 2 x^{2} + 2 x + 1 - 365 = 365 - 365$ $=>2x^2+2x+1color(red)(-365)=365color(red)(-365)$

$\Rightarrow 2 x^{2} + 2 x - 364 = 0$ $=>2x^2+2x-364=0$

Apply Quadratic Formula, that is $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} .$ $x=(-b+-sqrt(b^2-4ac))/(2a).$ Here

$a = 2, b = 2, c = - 364$ $a=2, b=2, c=-364$

$\Rightarrow x = \frac{- 2 \pm \sqrt{2^{2} - (4 \cdot 2 \cdot (- 364))}}{2 \cdot 2}$ $=>x=(-2+-sqrt(2^2-(4*2*(-364))))/(2*2)$

$\Rightarrow x = \frac{- 2 \pm \sqrt{4 - (8 \cdot (- 364))}}{4}$ $=>x=(-2+-sqrt(4-(8*(-364))))/(4)$

$\Rightarrow x = \frac{- 2 \pm \sqrt{4 - (- 2912)}}{4}$ $=>x=(-2+-sqrt(4-(-2912)))/4$

$\Rightarrow x = \frac{- 2 \pm \sqrt{4 + 2912}}{4}$ $=>x=(-2+-sqrt(4+2912))/4$

$\Rightarrow x = \frac{- 2 \pm \sqrt{2916}}{4}$ $=>x=(-2+-sqrt(2916))/4$

$\Rightarrow x = \frac{- 2 + 54}{4}$ $=>x=(-2+54)/4$ $or$ $or$ $x = \frac{- 2 - 54}{4}$ $x=(-2-54)/4$

$\Rightarrow x = \frac{52}{4}$ $=>x=52/4$ $or$ $or$ $x = \frac{- 56}{4}$ $x=(-56)/4$

$\Rightarrow x = 13$ $=>x=13$ $or$ $or$ $x = - 14$ $x=-14$

$x$ $x$ won't be negative because the value must be a positive integer.

So, $x = 13$ $x=13$

Then $x + 1 = 14$ $x+1=14$

So the positive consecutive integers whose sum of squares is 365 are $13$ $13$ $&$ $&$ $14 .$ $14.$

$\Rightarrow 13^{2} + 14^{2} = 365$ $=>13^2+14^2=365$

$\Rightarrow 169 + 196 = 365$ $=>169+196=365$

$\Rightarrow 365 = 365$ $=>365=365$

Answer 2 · 2017-08-12T18:40:19

The positive integers are $13 and 14$ $13 and 14$

Explanation:

Let the two integers be $x and (x + 1)$ $x and (x+1)$

The sum of their squares must be $365$ $365$

$x^{2} + {(x + 1)}^{2} = 365$ $x^2 + (x+1)^2 = 365$

$x^{2} + x^{2} + 2 x + 1 = 365 \leftarrow$ $x^2 +x^2 +2x+1 =365" "larr$ make the quadratic equal to $0$ $0$

$2 x^{2} + 2 x - 364 = 0 \leftarrow \div 2$ $2x^2 +2x-364=0" "larr div 2$

$x^{2} + x - 182 = 0$ $x^2 +x-182 =0$

As we are looking for two integers, the expression will factorise.

Factors which differ by $1$ $1$ are on either side of the square root.

$\sqrt{182} = 13.49$ $sqrt182 =13.49$

The likely factors are $13 and 14,$ $13 and 14,$ check: $13 \times 14 = 182$ $" "13 xx 14 = 182$

$(x - 13) (x + 14) = 0$ $(x-13)(x+14)=0$

Solving gives $x = 13 or x = - 14$ $x = 13 or x= -14$ (which we reject as negative.)

$x = 13 and x + 1 = 14$ $x =13 and x+1 = 14$

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