solve the polynomial y=2x^2-4x+1
I am sure neither your text nor your teacher asked you to "solve the polynomial y=2x^2-4x+1 "
are you graphing the function, or are you solving the quadratic y=2x^2-4x+1 ?
To solve the quadratic equation y = 2x^2 - 4x + 1, we can use the quadratic formula or factorization method. Let's solve it using both methods.
Method 1: Quadratic Formula
The quadratic formula states that for any quadratic equation ax^2 + bx + c = 0, the solutions for x can be found using the following formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Comparing this with our quadratic equation y = 2x^2 - 4x + 1, we can see that a = 2, b = -4, and c = 1.
Now we can substitute these values into the quadratic formula and solve for x:
x = (-(-4) ± √((-4)^2 - 4(2)(1))) / (2(2))
x = (4 ± √(16 - 8)) / 4
x = (4 ± √8) / 4
x = (4 ± 2√2) / 4
Simplifying further:
x = (2 ± √2) / 2
So the solutions to the quadratic equation y = 2x^2 - 4x + 1 are x = (2 + √2) / 2 and x = (2 - √2) / 2.
Method 2: Factorization
To solve the quadratic equation y = 2x^2 - 4x + 1 by factorization, we need to find two binomials whose product equals the quadratic equation.
First, let's try to find two numbers whose product is ac (in this case, a = 2 and c = 1) and whose sum is b (in this case, b = -4). We have:
ac = 2 * 1 = 2
b = -4
Now, we need to find two numbers that fit these criteria. The numbers are -1 and -2:
-1 * -2 = 2
-1 + (-2) = -3
Now, we can rewrite the quadratic equation y = 2x^2 - 4x + 1 in the factored form:
y = 2x^2 - 2x - 2x + 1
= 2x(x - 1) - (x - 1)
= (2x - 1)(x - 1)
Setting each factor equal to zero:
2x - 1 = 0 or x - 1 = 0
Solving for x:
2x - 1 = 0
2x = 1
x = 1/2
x - 1 = 0
x = 1
So the solutions to the quadratic equation y = 2x^2 - 4x + 1 are x = 1/2 and x = 1.
Note: In both methods, we have found the values of x that satisfy the given quadratic equation.