Math, asked by nayefsiddiqueoxgvw6, 11 months ago

find two consecutive odd integers sum of whose square is 970​

Answers

Answered by bhushanbathe
3

Step-by-step explanation:

may it will help u ...................

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Answered by sharonr
0

The consecutive odd integer are -23 and -21 or 21 and 23

Solution:

Let the two consecutive odd integers be "x' and x + 2

Given that,

Sum of squares is 970

Therefore,

x^2 + (x + 2)^2 = 970\\\\Expand\ using\ (a + b)^2 = a^2 + 2ab + b^2\\\\x^2 + x^2 + 4x + 4 = 970\\\\2x^2 + 4x + 4 - 970 = 0\\\\2x^2 + 4x - 966 = 0

Divide equation by 2

x^2 + 2x - 483 = 0\\\\Split\ the\ middle\ term\\\\x^2 + 23x - 21x - 483 = 0\\\\Group\ the\ expression\\\\(x^2 + 23x) - (21x + 483) = 0\\\\Factor\ the\ common\ term\ out\\\\x(x + 23) - 21(x + 23) = 0\\\\(x + 23)(x - 21) = 0

Thus,

x + 23 = 0

x = -23

And,

x - 21 = 0

x = 21

Thus the two consecutive odd integers are:

x = -23 and x + 2 = -23 + 2 = -21

The other consecutive odd integer is:

x = 21 and x + 2 =21 + 2 = 23

Thus the consecutive odd integer are -23 and -21 or 21 and 23

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