How do you find the roots, real and imaginary, of #y=x(2x-1)-(3x-1)^2 # using the quadratic formula?

Question

How do you find the roots, real and imaginary, of #y=x(2x-1)-(3x-1)^2      # using the quadratic formula?

Kennedy Adams · Accepted Answer

#x = 5/14+-(sqrt(3))/14i#

Let's multiply out the brackets and get ourselves some quadratic terms:

#y = 2x^2 - x - 9x^2 + 6x - 1 = -7x^2 +5x -1#

The quadratic formula for polynomial of the form #ax^2+bx+c = 0# is given by:

#x = (-b+-sqrt(b^2-4ac))/(2a)#

#a = -7, b = 5, c = -1#

#therefore x = (-5+-sqrt(25-4(-7)(-1)))/(2(-7))#

#x= (-5+-sqrt(-3))/(-14)#

Recall that #i^2 = -1# so we can rewrite #sqrt(-3)# as #sqrt(3i^2) = sqrt(3)i#

#therefore x = 5/14+-(sqrt(3))/14i#

Jameson Allen · Answer

undefined To find the roots of the equation $y = x(2x - 1) - (3x - 1)^2$ using the quadratic formula, first, rewrite the equation in the form $ax^2 + bx + c = 0$. Then, identify the values of $a$, $b$, and $c$. Finally, substitute these values into the quadratic formula $x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}$ to find the roots.

In this case:
- $a = 2$,
- $b = -1$,
- $c = -1$.

Substituting these values into the quadratic formula gives:
$$x = \frac{{-(-1) \pm \sqrt{{(-1)^2 - 4(2)(-1)}}}}{{2(2)}}$$

$$x = \frac{{1 \pm \sqrt{{1 + 8}}}}{4}$$

$$x = \frac{{1 \pm \sqrt{9}}}{4}$$

$$x = \frac{{1 \pm 3}}{4}$$

So the roots are:
$$x_1 = \frac{1 + 3}{4} = 1$$
$$x_2 = \frac{1 - 3}{4} = -\frac{1}{2}$$

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