Question

Given f(x)=x^2-3x+3, find the value(s) for x such that f(x)=31

Answer 1

To find the value(s) for x such that f(x) = 31 given f(x) = x^2 - 3x + 3, you set the equation equal to 31, solve for x using the quadratic formula, and find the two solutions x = 7 and x = -4.

Step-by-step explanation:

To find the value(s) for x such that f(x) = 31 when given f(x) = x^2 - 3x + 3, we need to set the function equal to 31 and solve for x:

x^2 - 3x + 3 = 31

Subtract 31 from both sides to set the equation to zero:

x^2 - 3x - 28 = 0

Now we use the quadratic formula to find the two possible values of x. The quadratic formula is:

x = (-b ± √(b^2 - 4ac))/(2a)

For the equation x^2 - 3x - 28 = 0, a = 1, b = -3, and c = -28. Plugging these values into the quadratic formula gives us:

x = (3 ± √(3^2 - 4(1)(-28)))/(2(1))

x = (3 ± √(9 + 112))/2

x = (3 ± √121)/2

x = (3 ± 11)/2

The two possible solutions for x are:

• x = (3 + 11)/2 = 14/2 = 7
• x = (3 - 11)/2 = -8/2 = -4

Therefore, the values for x such that f(x) = 31 are x = 7 and x = -4.

Answer 2

x=4 and x=7 when f(x)=31

Step-by-step explanation:

`f(x)=x^2-3x+3`

`31=x^2-3x+3`

`0=x^2-3x-28`

`0=(x-7)(x+4)`

`x=7` and
`x=4`

Therefore, the values for x such that f(x)=31 are 4 and 7.