Algebra 2 add quadratic function roots method
Add a video to the Algebra 2 course for finding the roots of a quadratic function.
f(x)=6x^2-7x+3
f(x)=6(x^2-7/6x+3/6)
f(x)=6(x^2-7/6x+1/2)
p=-7/6
q=1/2
(-p/2+u)(-p/2-u)=q
(7/12+u)(7/12-u)=1/2
(7/12)^2-u^2=1/2
49/144 - 1/2 = u^2
49/144 - 72/144 = u^2
-23/144 = u^2
u=i sqrt(23)/12
Roots: {7/12 + i sqrt(23)/12, 7/12 - i sqrt(23)/12}
f(x)=6x^2-7x+2
f(x)=6(x^2-7/6x+1/3)
p=-7/6, q=1/3
(7/12 - u)(7/12 + u)
49/144 - u^2 = 2/3
49/144 - 48/144 = u^2
u = sqrt(1/144) = 1/12
7/12 - 1/12, 7/12 + 1/12 = 6/12, 8/12
(x-6/12)(x-8/12)
x^2-6x/12-8x/12+48/144
x^2 - 14x/12 + 1/3
x^2 - 7/6x + 1/3
6(x^2-7/6x+1/3) = 6x^2-7x+2
Roots: {1/2, 1/3}
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