How would I complete the square with the following?
4x^2-25
I thought of $\displaystyle (4x+\frac{0}{2}^2)-25+\frac{0}{2}^2$ but that doesn't make any sense!
Any help please?
Why would you want to complete the square? This is the difference of two squares and will easily factor
$\displaystyle 4x^2-25 = (2x)^2 - 5^2$ and the difference of two squares is $\displaystyle a^2-b^2 = (a-b)(a+b)$
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If you want/must to complete the square you have the right idea but the first step is always to make sure the coefficent of $\displaystyle x^2$ is $\displaystyle 1$
$\displaystyle 4x^2-25 = x^2 - \dfrac{25}{4} = (x+0)^2 - \dfrac{25}{4} = (x+0)^2 - \left(\dfrac{5}{2}\right)^2$
You "complete the square" to get rid of a term of the form "ax", that is x to the 1 power, in a quadratic. In your example, $\displaystyle 4x^2- 25$, there is no "x" term- "$\displaystyle 4x^2$" is already a perfect square. The minimum value occurs when x= 0 and is -25. Since $\displaystyle 4x^2- 25= (2x- 5)(2x+ 5)$ the graph of $\displaystyle y= 4x^2- 25$ is a parabola with vertex at (0, -25), vertical axis, opening upward, and with x-intercepts at (5/2, 0) and (-5/2, 0).