Question

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

Answer 1

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

Explanation:

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

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

Explanation:

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]}