Question

Two years ago, father's age was 4 1/2 times son's age. Six years ago, , 2 father's age was twice the square of son's age. Find son's age today.​

Answer 1

Answer:

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Step-by-step explanation:

Let 2 years ago, son's age be x.

So my age(2 years ago) =4.5 x

So, 6 years ago,

My age=4.5x-4

Son's age=x-4

B/C,

4.5x-4=2((x-4)^2)

or, 4x^2–41x+72=0

Solving the quadratic, we get-

x=8 or x=2.5(not possible)

So my present age of=4.5x+2=38

My son's present age=x+2=10.(Ans)

Answer 2

$let \: m \: be \: the \: age \: of \: son$

$(m - 2) = \frac{9}{2} (s - 2)$

$2m - 4 = 9s - 18$

$2m = 9s - 14....(1)$

$6 \: \: years \: ago$

$(m - 6) = 2(s - 6) ^{2}$

$( \frac{9s - 14}{2} - 6) = 2(s - 6)^{2}$

$9s - 14 - 12 = 4( {s}^{2} - 12s + 36)$

$9s - 26 = 4 {s}^{2} - 48s + 144$

$4 {s}^{2} - 57s + 170 = 0$

$s = 57 \frac{ + }{ - } \sqrt{529} = \frac{57 \frac{ + }{ - } 23}{8} = 104.25$

$hence \: \: 4.25is \: the \: fraction$