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.

Guest Jun 24, 2021

#1**+3 **

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

\(x = {-(-8) \pm \sqrt{(-8)^2-4(33)(-1)} \over 2(33)}\)

ElectricPavlov Jun 24, 2021

#2**+2 **

$x^2+4x+4=100x^2-20x+1$

$x^2+4x+3=100x^2-20x$

$ 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

UsernameTooShort Jun 24, 2021

#5**+2 **

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!

UsernameTooShort
Jun 24, 2021

#3**+1 **

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

Guest Jun 24, 2021