What are the solutions of the quadratic equation?
2x^2 – 16x + 32 = 0
How do I solve this, I'm kinda new to quadratics and there confusing >.>
they're, not there.
first, factor out the 2 so you have smaller numbers to work with
2(x^2-8x+16) = 0
Now you want two numbers which add to -8 and multiply to +16
-4 and -4 fit the bill, so
2(x-4)(x-4) = 0
Now the answer is clear. x=4 is a double root.
To solve the quadratic equation 2x^2 – 16x + 32 = 0, you can use the quadratic formula or factorization method. Let's solve it using the quadratic formula. The quadratic formula is given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
Here, a, b, and c are the coefficients of x^2, x, and the constant term, respectively. In this equation, a = 2, b = -16, and c = 32.
Step 1: Identify the values of a, b, and c.
a = 2, b = -16, c = 32
Step 2: Substitute the values of a, b, and c into the quadratic formula.
x = (-(-16) ± √((-16)^2 - 4(2)(32))) / (2(2))
Simplifying further:
x = (16 ± √(256 - 256)) / 4
x = (16 ± √0) / 4
Step 3: Simplify and calculate the values of x.
x = (16 ± 0) / 4
Since the square root of zero is zero, we have:
x = 16 / 4
x = 4
The solution to the quadratic equation 2x^2 – 16x + 32 = 0 is x = 4.
So, the equation has one solution, x = 4.
To solve a quadratic equation, such as the one given: 2x^2 – 16x + 32 = 0, you can use the quadratic formula, because this equation is not factorable.
The quadratic formula states that for an 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 the equation 2x^2 – 16x + 32 = 0,
a = 2, b = -16, and c = 32.
Substituting these values into the quadratic formula, we get:
x = (-(-16) ± √((-16)^2 - 4(2)(32))) / (2(2))
x = (16 ± √(256 - 256)) / 4
x = (16 ± √0) / 4
x = (16 ± 0) / 4
x = 16 / 4
x = 4
So, the only solution to this quadratic equation is x = 4.