(2x - 1)² - (x - 1)²
Answers
Answer:
First, expand the exponents:
(2x+1)(2x+1) = (x+1)(x+1)
4x^2+4x+1 = x^2+2x+1
First, subtract all three terms on the right side from both sides:
3x^2+2x = 0
Factor out the x:
x(3x+2) = 0
This is a quadratic equation, and while it doesn’t look exactly like the typical form of a quadratic equation it does share one important attribute: it has two solutions.
You need to find two values for x that make this equation a true statement. The first value is zero 0:
x(3x+2) = 0
x = 0
0(3*0+2) = 0
0(2) = 0
0 = 0
The second value needs to be the opposite of positive 2; this will make the term inside of the parentheses equal to zero and also make this equation a true statement.
So, solve the parenthese for x:
3x+2 = 0
Subtract 2 from both sides:
3x = -2
Divide by 3:
x = -(2/3)
Now substitute for x to prove this:
x(3x+2) = 0
x = -(2/3)
-(2/3)*(3*-2/3+2) = 0
-(2/3)*(-6/3 + 2) = 0
-(2/3)*(-2+2) = 0
-(2/3)*0 = 0
0 = 0
Okay! So both of these values for x make this equation a true statement. So… are we done? Almost: we need to substitute for each value for x in the original equation!
(2x+1)^2 = (x+1)^2
x = 0
(0+1)^2 = (0+1)^2
1^2 = 1^2
1 = 1
So x = 0 is a valid solution for this equation. Now how about the other value?
(2x+1)^2 = (x+1)^2
x = -2/3
(2*-2/3 + 1)^2 = (-2/3 + 1)^2
Perform the multiplication on the left:
(-4/3 +1)^2 = (-2/3+1)^2
Convert all terms to like terms; the LCD is 3:
(-4/3 + 3/3)^2 = (-2/3 + 3/3)^2
(-1/3)^2 = (1/3)^2
Apply the exponents:
1/9 = 1/9
So x = -2/3 is also a valid solution to this equation. This means both of the solutions - x = 0 and x = -2/3 - are both capable of making this equation a true statement.
Answer:
using identity a² -b² =(a-b)(a+b)
{2x-1+x-1}{2x-1-x+1}
{3x-2}{x}
{3x² -2x}