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

Question

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

Eva Abramson · Accepted Answer

The roots are #x= 1/2# and #x=1/3#

We begin with #y=2x^2-x+(2x-1)^2#. Our first step is to simplify our equation until it cannot be reduced any further. Then we move on to the quadratic formula.

#y=2x^2-x+(2x-1)^2# can be changed to #y=2x^2-x+((2x-1)*(2x+1))#. That becomes #y=2x^2-x+4x^2-2x-2x+1# or #y=2x^2-x+4x^2-4x+1#. Now we combine like terms like #2x^2+4x^2# and #-x-5x#. That gives us #y=6x^2-5x+1#.

Now we can move on to the quadratic equation, which is #(-b+-sqrt(b^2-4*a*c))/(2a)# where those variables come from #ax^2+bx+c#. In our case #a=6#, #b=-5#, and #c=1#. Now we just plug in our value to the formulla: #(5+-sqrt(-5^2-4*6*1))/(2*6)#. This can be simplified to #(5+-sqrt(25-24))/(12)# or #(5+-1)/12#. That becomes #4/12# or #6/12#, which can be sipmlified to #1/3# and #1/2#, respectivley. Those are our roots, and we can graph the original equation and look at the #x#-intercepts (roots).

graph{y=6x^2-5x+1}

And they are! Nice work!

Savannah Anderson · Answer

undefined To find the roots of the quadratic equation $y = 2x^2 - x + (2x - 1)^2$, we first simplify it:

$$ y = 2x^2 - x + (4x^2 - 4x + 1) $$
$$ y = 2x^2 - x + 4x^2 - 4x + 1 $$
$$ y = 6x^2 - 5x + 1 $$

Now, we can use the quadratic formula to find the roots of this equation, where the quadratic formula is given by:

$$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$

For our equation $y = 6x^2 - 5x + 1$, the coefficients are $a = 6$, $b = -5$, and $c = 1$. Substituting these values into the quadratic formula, we get:

$$ x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \cdot 6 \cdot 1}}{2 \cdot 6} $$
$$ x = \frac{5 \pm \sqrt{25 - 24}}{12} $$
$$ x = \frac{5 \pm \sqrt{1}}{12} $$
$$ x = \frac{5 \pm 1}{12} $$

So, the roots are:

$$ x_1 = \frac{5 + 1}{12} = \frac{6}{12} = \frac{1}{2} $$
$$ x_2 = \frac{5 - 1}{12} = \frac{4}{12} = \frac{1}{3} $$

Therefore, the real roots of the equation are $x = \frac{1}{2}$ and $x = \frac{1}{3}$, and there are no imaginary roots.