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

1 Answer
Johnson Z.
Mar 16, 2016

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.

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