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two natural numbers differ by 4 and the sum of their squares is 58.Find the number.Plz fast
a sum of quadratic equation

Answer :

one number=x
another number=x+4

(x*x)+((x+4)(x+4))=58
x2+x2+8x+16=58
2xsquare+8x-42=0
xsquare+4x-21=0
xsquare+4x=21
xsquare+4x+2square=21+2square
(x+2)square=25
x+2=5
x=3

so numbers are 3 and 7

Let the smaller natural no. be ------>x , then
other natural no. would be -------> (x + 4)
According to the question,
x^2 + (x + 4)^2 = 58
x^2 + (x^2 + 8x + 16) = 58
2x^2 + 8x - 42 = 0 
x^2 + 7x - 3x - 21 = 0
x(x + 7) - 3(x + 7) = 0
(x + 7)(x - 3) = 0
now, either (x + 7) = 0 or (x - 3) = 0
if (x + 7) = 0 => x = -7 or  if (x - 3) = 0 => 3
x = 3
So, the required numbers are 3 & 7.

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