Question

(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

Answer 1

8x+x(x-6) or (x-8)^2=36 ?

Answer 2

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=?

Answer 3

a) take square root of both sides: x-8=pm 6
pm means plus or minus then add 8 on both sides

Answer 4

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

Answer 5

She said \(\pm 6\) which is not cancelling one of the roots.

Answer 6

\[(x-8)^2 = 36\]\[\implies x-8 = \pm6\]\[\implies x = 8\pm 6\]\[\implies x \in \{2,14\}\]

Answer 7

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!