Question

How do you find the roots, real and imaginary, of `y=2x^2 + 13x +6` using the quadratic formula?

Answer 1

real roots: `x=-1/2,-6`
imaginary roots: none

Explanation:

`1`. Since the given equation is already in standard form, identify the `color(blue)a,color(darkorange)b,` and `color(violet)c` values. Then plug the values into the quadratic formula to solve for the roots.

`y=color(blue)2x^2` `color(darkorange)(+13)x` `color(violet)(+6)`

`color(blue)(a=2)color(white)(XXXXX)color(darkorange)(b=13)color(white)(XXXXX)color(violet)(c=6)`

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

`x=(-(color(darkorange)(13))+-sqrt((color(darkorange)(13))^2-4(color(blue)2)(color(violet)(6))))/(2(color(blue)2))`

`x=(-13+-sqrt(169-48))/4`

`x=(-13+-sqrt(121))/4`

`x=(-13+-11)/4`

`x=(-13+11)/4color(white)(X),color(white)(X)(-13-11)/4`

`x=-2/4color(white)(X),color(white)(X)-24/4`

`color(green)(|bar(ul(color(white)(a/a)x=-1/2,-6color(white)(a/a)|)))`

`:.`, the real roots are `x=-1/2` or `x=-6`.