Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
pokemon23
Group Title
I need help with my math hw solving quadratic equations can someone help me understand it by helping guide me through I will surely appreciated the effort I'm here to learn :D
 2 years ago
 2 years ago
pokemon23 Group Title
I need help with my math hw solving quadratic equations can someone help me understand it by helping guide me through I will surely appreciated the effort I'm here to learn :D
 2 years ago
 2 years ago

This Question is Closed

pokemon23 Group TitleBest ResponseYou've already chosen the best response.1
The first question is \[x ^{2} 3x +2=0\]
 2 years ago

sonicx28@sbcglobal.net Group TitleBest ResponseYou've already chosen the best response.0
You can either inspect the quadratic equation to see what 2 numbers multiply to be +2 and add up to be 3. Then you can write rewrite the quadratic in terms of two factors (x2)(x1) = 0 then by the zero product property if one of the factors is zero then the whole thing is zero so you can then solve the equations: x 2 =0 and x1 = 0 and soon enough you'll find that x=2 and x=1 are the roots to the quadtratic.
 2 years ago

sonicx28@sbcglobal.net Group TitleBest ResponseYou've already chosen the best response.0
Another approach is to use the quadratic formula which is a surefire way of finding the roots but the catch is that it must be in the standard form which is \[y = ax^2 + bx + c\] and then you would use the quadratic formula: \[x = {b \pm \sqrt(b^24ac) \over 2a}\] the \[\sqrt(b^24ac)\] term is called the "discriminant" because if you get a negative number in that part of the equation then there are no real solutions and thus "discriminates" between real and imaginary solutions. Note that a quadratic equation has 2 roots...either both real or one real and one imaginary.
 2 years ago

marshallinwashington Group TitleBest ResponseYou've already chosen the best response.1
Sonic, I believe imaginary numbers come in pairs. So you have either two real roots or two imaginary roots.
 2 years ago

sonicx28@sbcglobal.net Group TitleBest ResponseYou've already chosen the best response.0
Another method you can use to solve quadratic equations is called "completing the square" which requires the coefficient of the \[x ^2\] term to be one. The basic process involves you taking the b coefficient in \[x^2 +bx+c\] and then dividing by 2 it and then squaring it and then plugging it back into the equation which will give us:\[(x^2 + bx +({b \over 2})^2 )  ({b \over 2})^2 + c = 0\]. You put that \[({b \over 2})^2\] to maintain equality in the equation. It then follows that The first part in parenthesis is \[x^2 + bx + ({b \over 2})^2 = (x+ ({b \over 2}))^2\]. You can now have\[(x + ({b \over 2}))^2 ({b \over 2})^2 + c = 0\] Then you would find the roots by solving for x by moving all the terms other than the perfect square to the other side then taking the square root of both sides and finally solving for x  remember when you take the square root of something you have to put a plus minus there. I hope that was of help.
 2 years ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.