How do you find the roots, real and imaginary, of #y=2x^2 + 13x + 6+4(x -1)^2 # using the quadratic formula?

Question

How do you find the roots, real and imaginary, of #y=2x^2 + 13x + 6+4(x -1)^2        # using the quadratic formula?

Avery Alexander · Accepted Answer

#x = (-5 + sqrt(-215))/12# or #x = (-5 - sqrt(-215))/12# First expand and simplify to ensure just one term for each power of x. #y = 2x^2 +13x +6 +4(x^2 -2x + 1)# #y = 2x^2 +13x + 6 + 4x^2 -8x +4# #y=6x^2 +5x +10# Then use #x = (-b +- sqrt(b^2 - 4ac))/(2a)# #x = (-5 +- sqrt( 5^2 - 4*6*10))/(2*6)# #x = (-5 +-sqrt(25 - 240))/12# #x = (-5 +- sqrt(-215))/12# There are only imaginary roots to this equation, meaning that in graphical terms the graph never intersects the x axis.

Jameson Allen · Answer

undefined To find the roots of the given equation, first, rewrite it in standard form: $y = 2x^2 + 13x + 6 + 4(x - 1)^2$

Expand the squared term: $y = 2x^2 + 13x + 6 + 4(x^2 - 2x + 1)$

Simplify: $y = 2x^2 + 13x + 6 + 4x^2 - 8x + 4$

Combine like terms: $y = 6x^2 + 5x + 10$

Now, the equation is in the form $ax^2 + bx + c$. To find the roots using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 6$, $b = 5$, and $c = 10$:

Substitute the values: $x = \frac{-5 \pm \sqrt{5^2 - 4*6*10}}{2*6}$

Calculate the discriminant: $5^2 - 4*6*10 = 25 - 240 = -215$

Since the discriminant is negative, the roots are complex.

Calculate the roots: $x = \frac{-5 \pm \sqrt{-215}}{12}$

Simplify: $x = \frac{-5 \pm i\sqrt{215}}{12}$

So, the roots are: $x = \frac{-5 + i\sqrt{215}}{12}$ and $x = \frac{-5 - i\sqrt{215}}{12}$.