# Answer to Question #16970 in Algebra for bluerose

Question #16970

word problem involving quadratic equation

mich is 10 times older than her daughter anne. in five years,the product of her age and anne's age was 175years.how old are each at present?

mich is 10 times older than her daughter anne. in five years,the product of her age and anne's age was 175years.how old are each at present?

Expert's answer

Mich is X years old, Anne is Y years old.Mich is 10 times older than her daughter Anne: X = 10 * Y.

In five years, the product of her age and Anne's age was 175 years:

(X + 5) * (Y + 5) = 175.

(10 * Y + 5) * (Y + 5) = 17510 * Y^2 + 5 * Y + 10 * Y * 5 + 5 * 5 = 175

10 * Y^2 + 55 * Y + 25 = 175

2 * Y^2 + 11 * Y + 5 = 352 * Y^2 + 11 * Y - 30 = 0

Y = { -11 ± Sq Root ( 121 + 240 ) } / 4 = {2, - 7.5}

At present, Anne is 2 years old and Mich is 10 * 2 = 20 years old.

In five years, the product of her age and Anne's age was 175 years:

(X + 5) * (Y + 5) = 175.

(10 * Y + 5) * (Y + 5) = 17510 * Y^2 + 5 * Y + 10 * Y * 5 + 5 * 5 = 175

10 * Y^2 + 55 * Y + 25 = 175

2 * Y^2 + 11 * Y + 5 = 352 * Y^2 + 11 * Y - 30 = 0

Y = { -11 ± Sq Root ( 121 + 240 ) } / 4 = {2, - 7.5}

At present, Anne is 2 years old and Mich is 10 * 2 = 20 years old.

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