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pokemon23
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
The first question is \[x ^{2} -3x +2=0\]
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 (x-2)(x-1) = 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 x-1 = 0 and soon enough you'll find that x=2 and x=1 are the roots to the quadtratic.
Another approach is to use the quadratic formula which is a sure-fire 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: \[x = {-b \pm \sqrt(b^2-4ac) \over 2a}\] the \[\sqrt(b^2-4ac)\] 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.
Sonic, I believe imaginary numbers come in pairs. So you have either two real roots or two imaginary roots.
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.