Find all real solutions to
x^2 + 4x + 4 = 100x^2 - 20x + 1
If you find more than one, then list the values separated by commas.
+36916 re-arrange to :
99x^2 -24x-3 = 0 Use quadratic formula
Simplify a bit by dividing by 3
33x^2 - 8x -1=0
a = 33 b = -8 c = -1
\(x = {-(-8) \pm \sqrt{(-8)^2-4(33)(-1)} \over 2(33)}\)
+171 $x^2+4x+4=100x^2-20x+1$
$x^2+4x+3=100x^2-20x$
$ x^2+24x+3=100x^2 $
$ -99x^2+24x+3=0 $
you can chose many ways to solve this, but i will do quadratic formula $ _2 x_1=\frac{-b\pm \sqrt{b^2-4ac}}{2a} $ where $ a=-99 \ ; \ b=24 \ ; \ c=3 $
$ _2 x_1=\frac{-24\pm\sqrt{576+1188} }{2\left(-99\right)} $
$ _2 x_1=\frac{-24\pm \:42}{2\left(-99\right)} $
$_2 x_1=\frac{-24\pm \sqrt{24^2-4\left(-99\right)\cdot \:3}}{2\left(-99\right)} $
$ x_1= \frac{18}{-198} $ and $ x_2 \frac{66}{198} $
which you can simpify:
$ x_1=\frac{-1}{11} $ and $x_2= \frac{1}{3} $
the reply above had a little mistakes :D
+36916 UTS.... there are no mistakes in my answer that I can find.....
+171 you are absolutely right -- i just perceived like something was off at the moment, dont know why -- wanted to edit the answer, and take the last line off, but it wouldn't let me!
Solution without the use of the quadratic formula:
Combine all terms to the right side of the equation:
99x^2 - 24x - 3 = 0
Divide by the common factor of 3:
33x^2 - 8X - 1 = 0
Factor the equation into two linear terms:
(11x + 1)(3x - 1) = 0
Since the product is zero, one or the other of the terms must be zero. Setting each to zero and solving for x gives the two solutions:
11x + 1 = 0
11x = -1
x = -1/11
3x - 1 = 0
3x = 1
x = 1/3
The two solution (both real) are: -1/11, 1/3
