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pokemon23
 3 years ago
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
pokemon23
 3 years ago
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

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

sonicx28@sbcglobal.net
 3 years ago
Best ResponseYou've already chosen the best response.0You 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.

sonicx28@sbcglobal.net
 3 years ago
Best ResponseYou've already chosen the best response.0Another 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.

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

sonicx28@sbcglobal.net
 3 years ago
Best ResponseYou've already chosen the best response.0Another 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.
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