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(x-8)^2=36 find exact and approximate solutions to three decimal places. a) solve 8x+x(x-6)=0 b) find the x-intercepts of f(x)=8x+x(x-6) a)what are the solution x=? c) what are the x-intercepts
8x+x(x-6) or (x-8)^2=36 ?
a) (x-8)^2=36 find exact and approximate solutions to three decimal places. b) solve 8x+x(x-6)=0 c) find the x-intercepts of f(x)=8x+x(x-6) 1)what are the solution x=?
a) take square root of both sides: x-8=pm 6 pm means plus or minus then add 8 on both sides
x^2-16X+64=36 X^2-16X+28=0 Then, just find it using the quadratic formula, so the answers will be: (16+/-12)/2= 14 or 2 Myiniaya i think that way is incorrect because you are cancelling one of the roots, and you need all the answers
She said \(\pm 6\) which is not cancelling one of the roots.
\[(x-8)^2 = 36\]\[\implies x-8 = \pm6\]\[\implies x = 8\pm 6\]\[\implies x \in \{2,14\}\]
for b) you have (8x+x)(x-6)=0 expand: 8x^2-48x+x^2-6x=0 simplify: 9x^2-54x=0 x=0 OR x= 5 and polpak, sorry didnt pay attention to the pm, so pm is +/-, and yes that would be correct!