find two parts of 40 such that the sum of their squares is 850
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let the two parts = x, 40-x
x^2 + (40-x)^2 = 850
x^2 + 1600 -80x +x^2 = 850
2x^2 -80x +1600 = 850
x^2 -40x + 800 = 425
x^2 -40x +375 = 0
solve the quadratic equation, you'll get x, the first number. the second number is 40-x
x^2 + (40-x)^2 = 850
x^2 + 1600 -80x +x^2 = 850
2x^2 -80x +1600 = 850
x^2 -40x + 800 = 425
x^2 -40x +375 = 0
solve the quadratic equation, you'll get x, the first number. the second number is 40-x
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25 and 15 are the answers
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