Find the x-intercepts (if any) for the graph of the quadratic function. 6x2 +12x+5=0
Give your answers in exact form. Show your work.
by precalc you should surely know how to find the roots of a quadratic. What did you get?
6x2 ?
ever hear of the quadratic formula?
x = (-12±√(144-120))/(2*6)
B^2-4AC=
144-120= 24
x=12+sqrt24 / 12
2*2*2*3
sqrt24=sqrt2*2*2*3= + 2* sqrt6
x=(12+2*2.449)/12
x =(12+√24)/12=1+1/6√ 6 = 1.408
you are correct about
x = (12+√24)/12
There is also
x = (12-√24)/12
To find the x-intercepts of a quadratic function in the form ax^2 + bx + c = 0, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac))/(2a)
In the given quadratic function, we have a = 6, b = 12, and c = 5.
Plugging these values into the quadratic formula, we get:
x = (-12 ± √(12^2 - 4 * 6 * 5))/(2 * 6)
Simplifying further:
x = (-12 ± √(144 - 120))/(12)
x = (-12 ± √(24))/(12)
x = (-12 ± 2√6)/(12)
Next, we simplify the expression by factoring out a common factor:
x = (1/6)(-12 ± 2√6)
Now we have the x-intercepts in exact form:
x = (-2 ± √6)/6
Thus, the x-intercepts (if any) for the given quadratic function are (-2 + √6)/6 and (-2 - √6)/6.