Question
If $\left(x-\frac{1}{2}\right)^2 - \left(x-\frac{3}{2}\right)^2 = x+2$, then the value of $x$ is:
The correct answer is
4
Solving the Algebraic Equation for x
The problem asks for the value of $x$ in the equation: $ \left(x-\frac{1}{2}\right)^2 - \left(x-\frac{3}{2}\right)^2 = x+2 $ We can simplify the left side of the equation using the difference of squares formula, $a^2 - b^2 = (a-b)(a+b)$.
Applying the Difference of Squares Formula
Let $a = \left(x-\frac{1}{2}\right)$ and $b = \left(x-\frac{3}{2}\right)$.
- Calculate $a-b$: $ a-b = \left(x-\frac{1}{2}\right) - \left(x-\frac{3}{2}\right) = x - \frac{1}{2} - x + \frac{3}{2} = \frac{2}{2} = 1 $
- Calculate $a+b$: $ a+b = \left(x-\frac{1}{2}\right) + \left(x-\frac{3}{2}\right) = 2x - \left(\frac{1}{2} + \frac{3}{2}\right) = 2x - \frac{4}{2} = 2x - 2 $
Now substitute these back into the difference of squares formula:
$ a^2 - b^2 = (a-b)(a+b) = (1)(2x-2) = 2x-2 $Solving for x
Equate the simplified left side to the right side of the original equation:
$ 2x-2 = x+2 $Rearrange the terms to solve for $x$:
- Subtract $x$ from both sides: $ 2x - x - 2 = x - x + 2 $ $ x - 2 = 2 $
- Add 2 to both sides: $ x - 2 + 2 = 2 + 2 $ $ x = 4 $
Thus, the value of $x$ is 4.
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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
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