Question

12. The two positive consecutive numbers whose squares have the sum 85 are​

Answer 1

Answer:

Let two consecutive positive no is x & x+1 .

By condition

X^2+(x+1)^2=85

X^2+x^2+1+2x=85

2x^2+2x+1=85

2x^2+2x=84

X^2+x=42

X^2+x-42=0

X^2+7x-6x-42=0

X(x+7)-6(x+7)=0

(X+7)(x-6)=0

X=-7 & x=6

Taking only postive no

X=6 &x=7

Answer 2

Answer:

Let the two consecutive natural numbers be "x" and "x+1"

$\implies$ Given that sum of their squares is 85.

Then by hypothesis, we get

$\tt{x {}^{2} + (x + 1) {}^{2} = 85}$

$\tt{x {}^{2} + x {}^{2} + 2x + 1 = 85}$

$\tt{2x {}^{2} + 2x + 1 - 85 = 0}$

$\tt{2x {}^{2} + 2x + 84 = 0}$

$\tt{x {}^{2} + 7x - 6x - 42 = 0}$

$\tt{x(x + 7) - 6(x + 7) = 0}$

$\tt{(x - 6) \: (x + 7) = 0}$

$\tt{x = 6 \: or \: x = 7}$

Case 1) If x = 6x + 1 = 6+1 = 7

Case 2) If x = 7x + 1 = -7 + 1 = -6

The consecutive numbers that the sum of squares is 85 are 6,7 and -6 , -7

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