Question

three consecutive numbers are such that the sum of squares of first two number is greater than the square of third by 65. find the number​

Answer 1

$\huge\mathcal{\fcolorbox{aqua}{azure}{\red{❖αηsωεя}}}$

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❏Let the numbers be (2n+1),(2n+3) and (2n+5).

According to the Question,

(sum of the squares of the first two numbers) = (square of the third number) + 65

=(2n+1)² + (2n+3)² = (2n+5)² + 65

=(2n)²+2(2n)(1)+(1)² + (2n)²+2(2n)(3)+(3)² = (2n)²+2(2n)(5n)+(5)²+65

=4n²+4n+1 + 4n²+12n+9 = 4n²+20n+25+65

=8n²+16n+10 = 4n²+20n+90

=4n²-4n-80 = 0

=4(n²-n-20) = 0

=n²-n-20 = 0

=n²-5n+4n-20 = 0

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

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

=n=5 or n=-4

Putting the values of n=5,

=2n+1=2(5)+1=11

=2n+3=2(5)+3=13

=2n+5=2(5)+5=15

Putting the values of n=-4,

=2n+1=2(-4)+1=-7

=2n+3=2(-4)+3=-5

=2n+5=2(-4)+5=-3

So, the numbers are either 11,13&15 or -7,-5&-3 respectively.

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@ Vaishnavi Sah_❤:)

Answer 2

$\huge\mathcal{\fcolorbox{aqua}{azure}{\red{❖αηsωεя}}}$

_______________...❀♡❀...______________

❏Let the numbers be (2n+1),(2n+3) and (2n+5).

According to the Question,

(sum of the squares of the first two numbers) = (square of the third number) + 65

=(2n+1)² + (2n+3)² = (2n+5)² + 65

=(2n)²+2(2n)(1)+(1)² + (2n)²+2(2n)(3)+(3)² = (2n)²+2(2n)(5n)+(5)²+65

=4n²+4n+1 + 4n²+12n+9 = 4n²+20n+25+65

=8n²+16n+10 = 4n²+20n+90

=4n²-4n-80 = 0

=4(n²-n-20) = 0

=n²-n-20 = 0

=n²-5n+4n-20 = 0

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

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

=n=5 or n=-4

Putting the values of n=5,

=2n+1=2(5)+1=11

=2n+3=2(5)+3=13

=2n+5=2(5)+5=15

Putting the values of n=-4,

=2n+1=2(-4)+1=-7

=2n+3=2(-4)+3=-5

=2n+5=2(-4)+5=-3

So, the numbers are either 11,13&15 or -7,-5&-3 respectively.

_______________...❀♡❀...______________

@Shivu

@Shivu