Question

If the difference of the squares of two numbers is 45, the square of the smaller number is 4 times the larger number, then the numbers are
(a) 9, 6 or 9, -6
(b) 5, 6 or 5,-6
(c) 9, 5 or 9, -5
(d) None of these

Answer 1

Hint: Start by letting the numbers be x and y, where x>y. Use the statement that the difference of the squares of the numbers is 45 to form the first equation, and the statement that the square of the smaller number is 4 times the larger number to form the second equation. Solve the equations to get the answer.

Complete step-by-step answer:
Let us start the solution to the above question by letting the numbers to be x and y, where x>y.
Now, it is given that the difference of the squares of the smaller and the larger number is 45. So, if we represent this mathematically, we get
 $ {{x}^{2}}-{{y}^{2}}=45.........(i) $
The other statement given in the question is that the square of the smaller number is 4 times the larger number. So, if we write this in the form of an equation, we get
 $ {{y}^{2}}=4x.......(ii) $
Now, if we substitute $ {{y}^{2}} $ from equation (ii) in equation (i), we get
 $ {{x}^{2}}-4x=45 $
 $ \Rightarrow {{x}^{2}}-4x-45=0 $
Now, we can write 4x as (9-5)x. So, if we use this in our equation, we get
 $ {{x}^{2}}-\left( 9-5 \right)x-45=0 $
 $ \Rightarrow {{x}^{2}}-9x+5x-45=0 $
 $ \Rightarrow x\left( x-9 \right)+5\left( x-9 \right)=0 $
 $ \Rightarrow \left( x+5 \right)\left( x-9 \right)=0 $
So, for two values of x, i.e., x=-5 and x=9 the equation is satisfied.
Now, let us put x=-5 in equation (ii). On doing so, we get
 $ {{y}^{2}}=4x $
 $ \Rightarrow {{y}^{2}}=4\left( -5 \right) $
 $ \Rightarrow {{y}^{2}}=-20 $
As the square of a real number can never be negative so, x cannot have the value -5.
Now let us put x=9 in equation (ii). On doing so, we get
 $ {{y}^{2}}=4x $
 $ \Rightarrow {{y}^{2}}=4\times 9 $
 $ \Rightarrow {{y}^{2}}=36 $
Now we know that $ {{a}^{2}}=b $ implies $ a=\pm \sqrt{b} $ . So, our equation becomes:
 $ {{y}^{2}}=\pm \sqrt{36} $
We also know that 36 is the square of 6.
 $ {{y}^{2}}=\pm 6 $
So, the correct answer is “Option A”.

Note: It is important that you remember that $ {{a}^{2}}=b $ implies $ a=\pm \sqrt{b} $ . If you forget tis and by chance take $ {{a}^{2}}=b $ implies $ a=\sqrt{b} $ , there is a possibility that you miss some of the possible solutions. It is also important that you interpret the statement that the square of the smaller number is 4 times the larger number correctly and don’t make the mistake $ {{x}^{2}}=4y $ .