Title: Math algebra - Find the type & number of solutions for each equation Post by: nini07 on May 9, 2011 note that x2 means x squared.
find the type & number of solutions for each equation 59.x2-6x= -5 61, x2-245x=-144 63. 3x2-5x+3=0 Title: Math algebra - Find the type & number of solutions for each equation Post by: star on May 9, 2011 \({{x}^{2}}-6{x}=-5\)
Something like this? As you can tell, it's a quadratic! \({{x}^{2}}-6{x}+5=0\) Now you solve for x: \({{x}^{2}}-6{x}+5=0\) Or, you can factor out an \({x}\): \({x}({x}-6)=5 {x}=-5 {x}-6=-5 {x}=-5+6 {x}=1 \) So our \({x}\) values are: -5 and 1 I will work on the rest for you. I love this math module! (https://biology-forums.com/Themes/default/images/bbc/tex.gif) Genius Idea :-:) Title: Math algebra - Find the type & number of solutions for each equation Post by: star on May 9, 2011 Woops I made a mistake with the factoring in the previous question :-\
Title: Math algebra - Find the type & number of solutions for each equation Post by: bio_man on May 9, 2011 Yeah, I can see you're struggling, why don't you put it in vertex form and from there, solve for the x's?
Title: Math algebra - Find the type & number of solutions for each equation Post by: bio_man on May 9, 2011 Lets start with where you left off from, Star.
\(x2 - 6x = -5\) x2 - 6x + 5 = 0 You need to find factors of \(5\) that add up to \(6\)... Let's try \(5 * 1 = 5\), but when we add them, we get \(+6\). So that won't work. Since, it's not working, we have to convert it into vertex form by completing the square and then manually solving for \(x\). \(x2 - 6x + 5 = 0\) \(\frac{-6}{2}=(-3)^2=9 x2 - 6x + 9 - 9 + 5 = 0 (x-3)^2 - 4=0 \) Now solve for \(x\): \(\sqrt{4} = \sqrt{(x-3)^2}\) +/-\(2 = x-3\) if +2: 2 + 3 = \(x = 5\) if -2: -2 + 3 = \(x = 1\) Don't forget to write a "therefore" statement :) Title: Math algebra - Find the type & number of solutions for each equation Post by: star on May 9, 2011 Awesome, but we ended up getting the same answers ^-^, so I guess I was going it right afterall. Nini, are you ok with the other two or would u like me to answer them as well? :D
Title: Math algebra - Find the type & number of solutions for each equation Post by: bio_man on May 9, 2011 61. \(x^2-245x+144 = 0\)
\(x^2-245x +\)\(({\frac{245}{2})^{2}}\)\(-({\frac{245}{2})^{2}}\)\( +144 = 0\) \((x - \frac{245}{2})^2 - 14862.25 = 0 +{sqrt{\14862.25}} + \frac{245}{2}= x = 244.41 -{sqrt{\14862.25}} + \frac{245}{2}= x = 0.58917 \) Title: Math algebra - Find the type & number of solutions for each equation Post by: nini07 on May 9, 2011 SO FAR IM DOING GOOD WITH THE OTHERS I JUST NEED HELP WITH THE MORE ALGEBRA PART. THANK YOU
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