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Home » One year ago, a man was 8 times as old as his son. Now his age is equal to the square of his son’s age. Find their present ages.
One year ago, a man was 8 times as old as his son. Now his age is equal to the square of his son’s age. Find their present ages.
Exercise 6D | Maths | Selina | Solving (simple) Problems (Based on Quadratic Equations)
One year ago, a man was 8 times as old as his son. Now his age is equal to the square of his son’s age. Find their present ages.

ACCORDING TO QUES,

How about we think about the current age of the child to be x years.

Along these lines, the current age of the man

    \[=\text{ }x2\text{ }years\]

One year prior,

Child’s age

    \[=\text{ }\left( x\text{ }\text{ }1 \right)\]

 a long time

Man’s age

    \[=\text{ }\left( x2\text{ }\text{ }1 \right)\]

a long time

ques suggests, that one year prior; the man was 8 times as old as his child.

    \[\left( x2\text{ }\text{ }1 \right)\text{ }=\text{ }8\left( x\text{ }\text{ }1 \right)\]

    \[x2\text{ }\text{ }8x\text{ }\text{ }1\text{ }+\text{ }8\text{ }=\text{ }0\]

    \[x2\text{ }\text{ }8x\text{ }+\text{ }7\text{ }=\text{ }0\]

    \[\left( x\text{ }\text{ }7 \right)\text{ }\left( x\text{ }\text{ }1 \right)\text{ }=\text{ }0\]

    \[x\text{ }=\text{ }7,\text{ }1\]

At the point when

    \[x\text{ }=\text{ }1\]

, then, at that point,

    \[x2\text{ }=\text{ }1\]

, which is absurd as father’s age can’t be equivalent to child’s age.

hence,

    \[x\text{ }=\text{ }7\]

is taken

then,

The current period of child = x years

    \[=\text{ }7\text{ }years\]

And the current period of man = x2 years

    \[=\text{ }49\text{ }years~\]

i

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