How do you find the roots, real and imaginary, of #y=x^2 - (x-2)(2x-1) # using the quadratic formula?

Question

3370 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-24T11:28:45

#x=frac{5\pm\sqrt17}{2}#

Given quadratic function:

#y=x^2-(x-2)(2x-1)#

#y=x^2-2x^2+5x-2#

#y=-x^2+5x-2#

The roots of given function will be at #y=0# i.e.

#-x^2+5x-2=0#

#x^2-5x+2=0#

Now, using quadratic formula, the roots of above quadratic equation are given as

#x_{1, 2}=\frac{-(-5)\pm\sqrt{(-5)^2-4(1)(2)}}{2(1)}#

#=\frac{5\pm\sqrt17}{2}#

Thus, the roots of given quadratic function are real & distinct given as

#x=frac{5\pm\sqrt17}{2}#

Answer 2 · 2018-07-24T11:41:22

The solutions are #S={0.44, 4.56}#

The quadratic equation is

#y=x^2-(x-2)(2x-1)#

#=x^2-(2x^2-5x+2)#

#=x^2-2x^2+5x-2#

#=>#, #-x^2+5x-2=0#

#a=-1#

#b=5#

#c=-2#

The discriminant is

#Delta=b^2-4ac=(5)^2-4(-1)(-2)=25-8=17#

As #Delta>0#, there are #2# real roots

The solution is

#x=(-b+-sqrtDelta)/(2a)#

#=(-5+-sqrt17)/(-2)#

#x_1=0.438#

#x_2=4.56#

graph{-x^2+5x-2 [-10, 10, -5, 5]}