4
Finding Integer Difference from Sum and Product
Let the two integers be denoted by $x$ and $y$.
- Given the sum: $x + y = 26$
- Given the product: $x \times y = 165$
- We need to find the difference: $|x - y|$.
We can use the algebraic identity relating the sum, product, and difference of two numbers:
$ (x - y)^2 = (x + y)^2 - 4xy $
Substitute the given values into the identity:
$ (x - y)^2 = (26)^2 - 4(165) $
Calculate the terms:
$ (26)^2 = 676 $
$ 4 \times 165 = 660 $
Now, compute $(x - y)^2$:
$ (x - y)^2 = 676 - 660 $
$ (x - y)^2 = 16 $
To find the difference, take the square root of both sides:
$ x - y = \sqrt{16} $
$ x - y = \pm 4 $
The difference between the two integers is $4$.
Important Questions from Algebra
- For positive non-zero real variables $x$ and $y$, if
$ln\left(\frac{x+y}{2}\right) = \frac{1}{2} [ln\left(x\right) + ln\left(y\right)]$
then, the value of $\frac{x}{y} + \frac{y}{x}$ is Given $f(x, y) = x^2 - 2xy + y^2$
The complete contour of the equation $f(x, y) = 1$ is described by the option(s) ___.
- It is given that $x$ and $y$ are integers in the following equation:
$$(x + y - 7)^2 + (y + 3x - 13)^2 = 0$$
The value of $(x^3 + y^3)$ is ________ (in integer). - If $pqr \neq 0$ and $p^{-x} = \frac{1}{q}$, $q^{-y} = \frac{1}{r}$, $r^{-z} = \frac{1}{p}$, what is the value of the product $xyz$?
- Two points $(4, p)$ and $(0, q)$ lie on a straight line having a slope of $3/4$. The value of $(p – q)$ is
