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

Question

Nathan Axel · Accepted Answer

#x = 7/2+-sqrt(85)/2#

By roots, I will assume you mean zeros.

First expand out the quadratic, then combine terms to bring it into standard form:

#2x^2-x-(x+3)^2 = 2x^2-x-(x^2+6x+9)#

#color(white)(2x^2-x-(x+3)^2) = 2x^2-x-x^2-6x-9#

#color(white)(2x^2-x-(x+3)^2) = x^2-7x-9#

This is in the form:

#ax^2+bx+c#

with #a=1#, #b=-7# and #c=-9#

It has discriminant #Delta# given by the formula:

#Delta = b^2-4ac = (color(blue)(-7))^2-4(color(blue)(1))(color(blue)(-9)) = 49+36 = 85#

Since #Delta > 0#, this quadratic has two distinct real zeros, but since #85# is not a perfect square, those zeros are irrational.

We can use the quadratic formula to find them:

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

#color(white)(x) = (-b+-sqrt(Delta))/(2a)#

#color(white)(x) = (7+-sqrt(85))/2#

#color(white)(x) = 7/2+-sqrt(85)/2#

Abigail Allen · Answer

undefined To find the roots of the quadratic equation y = 2x^2 - x - (x + 3)^2 using the quadratic formula, we first need to rewrite the equation in the standard form ax^2 + bx + c = 0. Then, we can identify the values of a, b, and c, and plug them into the quadratic formula:

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

Given the equation $y = 2x^2 - x - (x + 3)^2$, we can expand the expression $(x + 3)^2$ and rewrite the equation in standard form:

$$y = 2x^2 - x - (x^2 + 6x + 9)$$
$$y = 2x^2 - x - x^2 - 6x - 9$$
$$y = x^2 - 7x - 9$$

Now, we can identify the coefficients a, b, and c:

$$a = 1$$
$$b = -7$$
$$c = -9$$

Next, plug these values into the quadratic formula:

$$x = \frac{{-(-7) \pm \sqrt{{(-7)^2 - 4(1)(-9)}}}}{{2(1)}}$$

Solve for the discriminant:

$$b^2 - 4ac = (-7)^2 - 4(1)(-9) = 49 + 36 = 85$$

$$x = \frac{{7 \pm \sqrt{85}}}{{2}}$$

So, the roots are:

$$x_1 = \frac{{7 + \sqrt{85}}}{{2}}$$

$$x_2 = \frac{{7 - \sqrt{85}}}{{2}}$$

These are the real and imaginary roots of the given quadratic equation.