I cannot rationalize why I come up with x =[ +/- b +/- (b^2-4ac)^(1/2)]/2a instead of simply x =[ b +/- (b^2-4ac)^(1/2)]/2a. Help appreciated, thank you,
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Neither is correct, not to mention that you need grouping symbols around your denominator.I cannot rationalize why I come up with x =[ +/- b +/- (b^2-4ac)^(1/2)]/2a instead of simply x =[ b +/- (b^2-4ac)^(1/2)]/2a. Help appreciated, thank you,
Neither is correct, not to mention that you need grouping symbols around your denominator.
It should be: x = [-b +/- (b^2 - 4ac)^(1/2)]/(2a).
No, sometimes it is the "divide symbol." Check the context.x = [-b +or- (b^2 - 4ac)^(1/2)] / (2a) \(\displaystyle \ \ \ \ \ \ \ \) The "/" is to go right up against the "]" and the "(,"
otherwise the intended numerator and denominator are just "floating" on opposite sides of the "/" symbol.
/ is divide symbol.
Sorry, I am not sure what statement you are pointing to. I tried to number most statements, can you be more specific please?-(-) is + not +/-
I've always thought that completing the square was the easiest way to derive the quadratic formula:
\(\displaystyle ax^2+bx+c=0\)
\(\displaystyle ax^2+bx=-c\)
\(\displaystyle x^2+\dfrac{b}{a}x=\dfrac{-c}{a}\)
\(\displaystyle \left (\dfrac{b}{2a} \right)^2=\dfrac{b^2}{4a^2}\)
\(\displaystyle x^2+\dfrac{b}{a}x+\dfrac{b^2}{4a^2}=\dfrac{-c}{a}+\dfrac{b^2}{4a^2}\)
\(\displaystyle \left (x+\dfrac{b}{2a} \right)^2=\dfrac{b^2-4ac}{4a^2}\)
\(\displaystyle x+\dfrac{b}{2a}=\pm \sqrt{\dfrac{b^2-4ac}{4a^2}}\)
\(\displaystyle x+\dfrac{b}{2a}=\dfrac{\pm \sqrt{b^2-4ac}}{2a}\)
\(\displaystyle x=\dfrac{-b}{2a} \pm \dfrac{\sqrt{b^2-4ac}}{2a}\)
\(\displaystyle x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}\)