Question

the difference of the squares of two natural numbers is 45.the square of the smaller number is four times the larger number.find the numbers

Answer 1

Let,
the smaller number is x
the larger number is y

Given,
y^2-x^2=45
The square of smaller number=4(larger number)
x^2=4y

y^2-4y=45
y^2-4y-45=0
a=1,b=-4,c=-45

y=(-b+-√b^2-4ac)/2a
y=(-(-4)+-√(-4)^2-4(1)(-45))/2(1)
y=(4+-√16+180)/2
y=(4+-√196)/2
y=(4+-14)/2
y=(4+14/2),(4-14/2)
y=9,-5
Here -5 is removed because it is negative.

x^2=4y
x=√4y
x=√4(9)
x=√36
x=6

Therefore,x=6
y=9.
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