Determine the solution set of (3x + 1)2 - 100 = 0.

Question

anonymous · Answer

its 3x +1 squared

anonymous · Answer

First decompose your (3x +1)^2.
it will yield:
$$(3x + 1)^2 - 100 = 0 \rightarrow 9x^2 + 6x + 1 - 100 = 0 \rightarrow 9x^2 + 6x - 99 = 0$$

Use the quadratic formula.
$$x = { -6 \pm \sqrt{36 + 4*9*99} \over 2*9 } = {-6 \pm \sqrt{3600} \over 18 } = -{1 \over 2} \pm  {60 \over 18} = - {3 \over 6} \pm  {8 \over 6}$$

So your solutions are x = -11/6 and x = 5/6

anonymous · Answer

$$(3x+1)^{2}-10^{2} =0$$
3x+1+10 =0  or 3x+1-10=0
3x+11=0 or 3x-9=0
3x=-11 or 3x=9
x=-11/3 or x=3

anonymous · Answer

@soati $$(-6\pm60)/18 =(-1\pm10)/3=-11/3 or 9/3 = -11/3 or 3$$

anonymous · Answer

6/18 should be 1/3 rather 1/2 as in your calculations. Apart from this you did it good

anonymous · Answer

Youre right, my bad. I was focused on writting all the { x over y }'s hehe.

Still my way works for a wider number of problems