Given f(x) = 3x^2 + 2x + 7. Find f(x)=0 (please explain steps)?

Question

Given f(x) = 3x^2 + 2x + 7. Find f(x)=0 (please explain steps)

Answer 1

go to calc101.com it will explain the steps for all derivatives free but you have to pay for a password to get the steps for integrals. This website is very helpful as a learning tool but I don't recommend using it to cheat in a class.

Answer 2

setting f(x)=0 is called finding the roots of a function. Another way to think about it is: If you make a graph of the function, f(x)=0 is where the graph touches the x axis. so:

0=3x^2 + 2x + 7
Now, if you can recognize the correct breakdown instantly, props. I, on the other hand, am going to use the quadratic equation: ax^2 +bx + c = [-b (+or-) sqrt(b^2 - 4*a*c)]/2a

[-2 (+or-) sqrt(4-4*3*7)]/2
=
[-2(+or-)sqrt(4-84)]/2

Now, if you look at this equation, you notice the square root of a negative number. This can mean only one thing: Nowhere on the graph does f(x) actually hit zero, which means that F(x) can never = zero. So, No solution.

Answer 3

1. Find the coefficients of the quadratic equation in the form 0=ax^2 + bx^1 + cx^0 In this case, a=3 b=2 c=7
2. Use coefficients in the quadratic equation: x=(-b±sqroot(b^2 - 4ac)) / 2a
3. Solve the equation:
x=(-2±sqroot(2^2 - 4*3*7)) / 2*3
x=(-2±sqroot(4-84)) / 6
x=(-2±sqroot(-80)) / 6

We can stop here. When the discriminant (b^2 - 4ac) is less than zero, it is impossible to evaluate the square root. In this case, the parabola described never crosses the x-axis, and f(x) never equals zero. Unless you are looking for imaginary solutions, of which there are 2. Here is how to find the imaginary solutions:

x=(-2±sqroot(80) * sqroot(-1)) / 6
x=(-2± sqroot(80) * i) / 6
x=(-2± 4 * sqroot(5) * i) / 6
x=(-2± 4i * sqroot(5)) / 6
x=(-1± 2i * sqroot(5)) / 3

Because of the plus or minus sign (± if you cant see it) there are two unique solutions to this parabola, albeit imaginary ones.

Answer 4

0 = 3x²+2x+7
Apply the quadratic equation
a= 3, b = 2, c = 7
(-b±√(b²-4ac) / (2a)
(-2±√(2²-(4)(3)(7)) / ((2)(3))

(-2+√(2²-(4)(3)(7)) / ((2)(3))
(-2+√(4-84) / 6
(-2+√(-80)) / 6
(-2+4√(-5)) / 6
-2+4√5i / 6

(-2-√(2²-(4)(3)(7)) / ((2)(3))
-2-√(80) / 6)
-2-4√5i / 6

There are no real roots. The two imaginary roots are
x = (-2+4√5i) / 6 and x = (-2-4√5i) / 6

Answer 5

Use the quadratic formula.
In this case, a = 3, b = 2, c = 7.

Your solutions for f(x) = 0 will be
(-b + sqrt(b^2 - 4ac))/(2a) and (-b - sqrt(b^2 - 4ac))/(2a).

In this case, there are no real solutions because b^2 - 4ac < 0.

Answer 6

0= 3x^2 + 2x + 7
You use quadratic formula
(-2+-sqrt(4-4(3)(7))) / 6
(-2+- sqrt -80) /6
x= -3 + (2isqrt(5)) / 3
x= -3 - (2isqrt(5)) / 3

Answer 7

3x^2 + 2x + 7=0
a=3
b=2
c=7
and solve for the roots using
x = (-b ± √(b²-4ac))/2a
And, just as a heads up, the roots are going to be complex valued ☺

HTH

Doug

Answer 8

f(x)=3x^2-2x replace each and every place there is an (x) with (a+a million)!!! it quite is incredibly undemanding. so 3x^2-2x will become 4a2++3a+a million which simplifies to 3a2+4a+a million that's comparable to your answer.