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The sum and product of two integers are $26$ and $165$ respectively. The difference between these two integers is ______

1. $2$
2. $3$
3. $4$
4. $6$

edited | 89 views
Migrated from GO Mechanical 6 months ago by Arjun

Let the two integers be $x$ and $y.$

Given that:

• $x+y=26\qquad \to (1)$
• $xy=165\qquad \to(2)$

To be find$:x-y=?$

We know that $(x-y)^{2}=x^{2}+y^{2}-2xy$

$\implies(x-y)^{2}=x^{2}+y^{2}+2xy-4xy$

$\implies(x-y)^{2}=(x+y)^{2}-4xy$

Put the values from equation $(1)$ and $(2)$ and we get

$\implies(x-y)^{2}=(26)^{2}-4\times165$

$\implies(x-y)^{2}=676-660$

$\implies(x-y)^{2}=16$

$\implies(x-y)=\sqrt{16}$

$\implies x-y=\pm4$

$\implies x-y=4$  $(or)$  $x-y=-4$

$-4$ is ruled out as both the sum and product of the numbers are positive meaning both the numbers must be positive.

So, $(C)$ is the correct choice.

by Veteran (54.8k points)
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we know (x-y)^2=(x+y)^2-4xy

x+y=26,xy=165

so (x-y)^2=26^2-4*165=16

so x-y=4
by (45 points)