the sum of the square of two consecutive natural number is 421 find the number
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Answered by
6
Let the 2 concecutive no.s be x and x+1
Sum of their squares is 421 (given)
i.e., x² + (x+1)² = 421
x²+x²+2x+1=421
2x²+2x=421-1
2(x²+x)=420
x²+x=210
x²+x-210=0
x²+15x-14x+210=0
x(x+15) -14(x+15)
(x+15)(x-14)=0
x+15=0 or x-14=9
x= -15 or x=14
-15 is not a natural number
So, First number x=14
and second number = x+1 = 14+1 =15
Hence the 2 natural numbers are 14,15
Answered by
26
Solution :
Let's 1st number = x
2nd number = x + 1
According to question :
x² + (x+1)² = 421
x² + x² + 2x + 1 = 421
2x² + 2x + 1 - 421 = 0
2x² + 2x - 420 = 0
x² + x - 210 = 0
x² + 15x - 14x - 210 = 0
x ( x + 15 ) -14 ( x + 15 ) = 0
( x - 14 ) ( x + 15 )
x - 14 = 0
x = 14
x + 15 = 0
x = - 15
-15 is not a natural number so value of x is 14
1st number = 14
2nd number = 14 + 1 = 15
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