aama100 4 years ago (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

1. imranmeah91

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

2. aama100

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=?

3. myininaya

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

4. saljudieh07

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

5. polpak

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

6. polpak

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

7. saljudieh07

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!