solve and check the function f(x)=3x^2+12x+3

Question

nothing else.

Answer 1

Let us assume f (x) = 3x ^2 + 12x + 3 = 0.

3x ^2 + 12x + 3 = 0

Divide each side by 3

x ^2 + 4x + 1 = 0.

x ^2 + 4x + 1 = 0 is quadratic, use the Quadratic Formula to find the roots of the related quadratic equation.

x = [- b ± sqrt(b ^2 - 4ac)] / 2a.

Substitute b = 4, a = 1, and c = 1.

x = [- 4 ± sqrt(4^2 - (4 * 1 * 1))] / 2 * 1

= [- 4 ± sqrt(16 - 4)] / 2

= [- 4 ± sqrt(12)] / 2

= [- 4 ± 2sqrt(3)] / 2

= - 2 ± sqrt(3).

Sulution of the function is x = - 2 ± sqrt(3).

Check :

Substitute the x = - 2 + sqrt(3) in 3x ^2 + 12x + 3 = 0.

3(- 2 + sqrt(3)) ^2 + 12(- 2 + sqrt(3)) + 3 = 0.

3(4 + 3 - 4sqrt(3)) - 24 + 12sqrt(3) + 3 = 0

3(7 - 4sqrt(3)) - 21 + 12sqrt(3) = 0

21 - 12sqrt(3) - 21 + 12sqrt(3) = 0

0 = 0

The above statement is true, so x = - 2 + sqrt(3) is a solution of 3x ^2 + 12x + 3 = 0.

Substitute the x = - 2 - sqrt(3) in 3x ^2 + 12x + 3 = 0.

3(- 2 - sqrt(3)) ^2 + 12(- 2 - sqrt(3)) + 3 = 0.

3(4 + 3+ 4sqrt(3)) - 24 - 12sqrt(3) + 3 = 0

3(7 + 4sqrt(3)) - 21 - 12sqrt(3) = 0

21 + 12sqrt(3) - 21 - 12sqrt(3) = 0

0 = 0

The above statement is true, so x = - 2 - sqrt(3) is a solution of 3x ^2 + 12x + 3 = 0.