solve the quadratic equation: x^2-4x+3=0
This equation is asking us to find the value(s) of x that when substituted in place of x in the equation x2-4x+3 the computed value should equal 0. There three potential methods of solving this equation, the first of which is factorisation: the best means of finding out the value of x would be to factorize the equation shown into two smaller linear equations that are equivalent to the quadratic; (x-1)(x-3) are these linear equations. Given that at least (x-1), or (x-3) must equal zero, 1 & 3, are the values of x that satisfy the equation.
Completing the square: this method on top of finding the roots can also be used to find the local extremum of the graph y=x2-4x+3. Ignoring the plus three, x2-4x is equivalent to (x-2)2-4 (the -4 is used to remove the additional constant 4 when expanding out the square), which when adding the remaining plus three, makes this square identical to the quadratic. From here (x-2)2-1=0, so, (x-2)2=1, and x-2=+-1, therefore x=2+-1. Thus giving the results x=1, x=3. Finally the quadratic formula: this is a given formula for calculating the quadratic while also identifying whether the quadratic is solvable. ax2+bx+c=0, can be identified by x=(-b+-(b2-4ac)1/2)/2a. If we assume a=1, b=-4,& c=3, then, x=(4+-((-4)2-4x1x3)1/2)/2x1=(4+-(4)1/2)/2=(4+-2)/2=1&3. Therefore the values of x are 1 and 3. If the value of the discriminant,(b2-4ac)>0 then the quadratic has 2 roots, if it equals 0 then there is one repeated root and if its less than 0 then the quadratic is unsolvable, as the square root of a negative number isn't real.
