solve the polynomial y=2x^2-4x+1
For Further Reading
algebra - Reiny, Thursday, May 17, 2007 at 10:34pm
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 ?
I am trying to
solve the quadratic sorry.
2x^2-4x+1 = 0 does not factor, so use the formula
x =(4 ± sqrt(16-4(2)(1)))/4
=(4 ± √8)/4
=(2 ± √2)/2
if you are looking for the vertex of
y=2x^2-4x+1
a quick way is to find the x of the vertex by x= -b/(2a) = 4/4 = 1
now sub x=1 back in your function to get the y of the vertex.
the other way would be to complete the square for 2x^2-4x+1
To make sure i did this right does y=1 also?
To solve the quadratic equation y=2x^2-4x+1, we can use the quadratic formula. The quadratic formula states that for any equation of the form ax^2+bx+c=0, the solutions for x can be found using the formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
For our equation y=2x^2-4x+1, the coefficients are a=2, b=-4, and c=1. Let's substitute these values into the formula:
x = (-(-4) ± √((-4)^2 - 4(2)(1))) / (2(2))
= (4 ± √(16 - 8)) / 4
= (4 ± √8) / 4
= (4 ± √(4 * 2)) / 4
= (4 ± 2√2) / 4
= 1 ± √2/2
Therefore, the solutions for x are x = 1 + √2/2 and x = 1 - √2/2.
Now, if you're looking for the vertex of the quadratic y=2x^2-4x+1, you can use the formula x = -b/(2a) to find the x-coordinate of the vertex. In this case, a=2 and b=-4, so:
x = -(-4)/(2(2))
= 4/4
= 1
To find the y-coordinate of the vertex, substitute the x-coordinate back into the equation:
y = 2(1)^2 - 4(1) + 1
= 2 - 4 + 1
= -1
Therefore, the vertex of the quadratic is (1, -1).
And to answer your question, no, y=1 is not a solution to the quadratic equation y=2x^2-4x+1. The equation y=1 represents a horizontal line, and the quadratic equation has two distinct solutions for x as mentioned earlier.