let f(x)=2x^2+7x+6 find all x such that f(x)=0
what is that
You are just solving the quadratic
2x^2+7x+6 = 0 , which just happens to factor nicely to
(x+2)(2x+3) = 0
so x = -2 or x = -3/2
To find all x such that f(x) = 0, we need to solve the quadratic equation 2x^2 + 7x + 6 = 0. Here's how you can find the solutions:
Step 1: Start with the quadratic equation in the form ax^2 + bx + c = 0.
In our case, a = 2, b = 7, and c = 6. So the equation becomes:
2x^2 + 7x + 6 = 0
Step 2: Use the quadratic formula to solve for x.
The quadratic formula states that for any quadratic equation ax^2 + bx + c = 0, the solutions for x are given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
In our equation, a = 2, b = 7, and c = 6. Substituting these values into the formula, we get:
x = (-7 ± √(7^2 - 4(2)(6))) / (2(2))
Simplifying further, we have:
x = (-7 ± √(49 - 48)) / 4
x = (-7 ± √1) / 4
Step 3: Simplify the square root.
√1 = 1, so we have:
x = (-7 ± 1) / 4
Step 4: Calculate the two possible solutions.
For the "+" case:
x = (-7 + 1) / 4
x = -6 / 4
x = -3/2
For the "-" case:
x = (-7 - 1) / 4
x = -8 / 4
x = -2
Therefore, the solutions to the equation 2x^2 + 7x + 6 = 0 are x = -3/2 and x = -2.