Question

Aastha multiplied two numbers and got -4. She then subtracted the numbers and got 5. Find the two numbers.

Answer 1

Hint: Assume that the two numbers are x and y. Using the fact that the product of the two number is -4 form an equation in x and y. Using the fact that the difference between x and y is 5, form another equation in x and y. Using the second equation eliminate the variable x from the first equation and hence form a quadratic equation in y. Solve for y and hence find the value of x and y and hence find the numbers.

Complete step by step solution:
Let the number be x and y.
Since the product of the numbers is -4, we have
$xy=-4\ \ \left( i \right)$
Also, since the difference of the number is 5, we have
$x-y=5\text{ }\left( ii \right)$
From equation (ii), we have
$x=y+5$
Substituting the value of x in equation (i), we get
$\begin{align}
  & \left( y+5 \right)y=-4 \\
 & \Rightarrow {{y}^{2}}+5y=-4 \\
\end{align}$
Adding 4 on both sides, we get
${{y}^{2}}+5y+4=0\text{ }\left( iii \right)$
We shall solve this quadratic equation using the method of splitting the middle term.
Since 5 = 4+1 and 4 = 4(1), we have
${{y}^{2}}+4y+y+4\times 1=0$
Taking y common from the first two terms and 1 common from the last two terms, we get
$y\left( y+4 \right)+1\left( y+4 \right)=0$
Taking y+4 common, we get
$\left( y+4 \right)\left( y+1 \right)=0$
Hence, we have
$y=-4$ or $y=-1$
If y = -4, from equation (ii), we have x = 5-4 = 1 and if y = -1, from equation (ii), we have x = 5-1 = 4
Hence the number or 4 and -1 or -4 and 1.


Note: Alternatively, we can solve the question as follows
We have
 $\begin{align}
  & {{\left( x+y \right)}^{2}}={{\left( x-y \right)}^{2}}+4xy={{5}^{2}}-16=9 \\
 & \Rightarrow x+y=\pm 3 \\
\end{align}$
We know that the quadratic equation with sum of roots as “a” and product of roots as b is ${{x}^{2}}-ax+b=0$
Hence, we have
$x$ and $y$ are the roots of the equation ${{t}^{2}}-3t-4=0$ or ${{t}^{2}}+3t-4=0$
From first equation, we have
 $\begin{align}
  & {{t}^{2}}-4t+t-4=0 \\
 & \Rightarrow \left( t-4 \right)\left( t+1 \right)=0 \\
 & \Rightarrow t=4,-1 \\
\end{align}$
From second equation, we have
$\begin{align}
  & {{t}^{2}}+3t-4=0 \\
 & \Rightarrow {{t}^{2}}+4t-t-4=0 \\
 & \Rightarrow \left( t+4 \right)\left( t-1 \right)=0 \\
 & \Rightarrow t=-4,1 \\
\end{align}$
Hence, we have
The numbers are 4 and -1 or -4 and 1.