Need help with quadratic formula, postive and negative signs? getting confused

P

Pencil

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I'm trying to learn the quadratic formula and I understand it to about 90%. My problem is, I don't understand when should b (ax^2 +bx + c =0) be negative or positive with regards to using the quadratic formula. I took a picture and posted the link at the bottom. I Would like to know what I'm doing wrong. I know X= +5 and it should really be -5 but that's where I'm stuck. Thanks in advance.

http://postimg.org/image/s7t58usxn/

Hope the link works, if not, I'll try to type it in manually within 24hrs.
 
Pencil said:
I'm trying to learn the quadratic formula and I understand it to about 90%. My problem is, I don't understand when should b (ax^2 +bx + c =0) be negative or positive with regards to using the quadratic formula. I took a picture and posted the link at the bottom. I Would like to know what I'm doing wrong. I know X= +5 and it should really be -5 but that's where I'm stuck. Thanks in advance.

http://postimg.org/image/s7t58usxn/

Hope the link works, if not, I'll try to type it in manually within 24hrs.
Click to expand...

It's
\(\displaystyle X_{1,2}=\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) see the difference ? So -b is always the opposite than that number next to X. You just change the plus minus sign in front of the square root. And all should be divided by 2a not just the square root like you wrote.
You wrote :
\(\displaystyle X_{1,2}=-b \pm \dfrac{\sqrt{b^2 - 4ac}}{2a} \)

So.....

\(\displaystyle x^2+7x+10=0 \\ ax^2+bx+c=0 \\ a=1 \qquad b=7\qquad c=10 \\ X_{1,2}=\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\ X_{1,2}=\dfrac{-(7) \pm \sqrt{(7)^2 - 4\cdot 1 \cdot 10}}{2\cdot 1} \\ X_{1,2}=\dfrac{-7 \pm \sqrt{9}}{2} \\ X_{1,2}=\dfrac{-7 \pm 3}{2} \\ X_{1}=\dfrac{-7 - 3}{2} \qquad X_{2}=\dfrac{-7 + 3}{2} \\ X_{1}=-5 \qquad X_{2}=-2 \)
 
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That was helpful. Thank you! I definitely need more practice with that formula :D
 
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