Question

If 3x^2 -x+9=0 what are the values of x?

Answer 1

Discriminant (Not important but in this case, if you want to know that if equations have real roots or complex.)

$D=b^2-4ac\\D=(-1)^2-4(3)(9)\\D=1-108\\D=-107$

Therefore, Our D<0, meaning that the equation has no real roots. And if you want to know the "real" values of x then there are none. Because if you draw the graph, you see that Parabola doesn't have any x-intercepts. (2 real roots when intercepts 2 x's, 1 real root when intercept only 1 x, and none when doesn't intercept any x's.)

You can solve the equation, but you don't get real roots answer, except you get complex solution.

$3x^2-x+9=0$

We can't find the values of x by factoring (Only when the equation has a complex solution.) So we are gonna use The Quadratic Formula.

$x=\frac{-b±\sqrt{b^2-4ac} }{2a}$ (Ignore the A^ after -b)

From the formula, a = 3, b = -1 and c = 9.

$x=\frac{-(-1)±\sqrt{(-1)^2-4(3)(9)} }{2(3)} \\x=\frac{1±\sqrt{-107}}{6}$

If you remember, negative numbers inside the square root don't exist in "Real Number" so we convert √-107 to √107 i

$x=\frac{1±\sqrt{107}i }{6}$

{i = √-1, i²=-1, i³= -i or -√-1 but because negative numbers in square root don't exist in "Real Number" so "-i" is better.}

i^4 = 1 and repeat again. i^5 = √-1, i^6 = -1 then again and again.

If you haven't learnt the Imaginary or Complex Number yet then I'd suggest you to answer "No real roots." or if you have and the equation here is part of Complex Number then answer x = 1±√107 i / 6. They both work.

Answer 2

Answer:

x = 1.89 or 1.56

Step-by-step explanation:

This can be solved by either factorisation Method or formulae method. But for the purpose of this question, formulae method will be used.

(-b±(√b²-4ac))/2a

3x^2 -x+9=0

Where

a=3

b=-1

C=9

(-(-1)±(√(-1²)-4(3)(9))/2(3)

(1±(√1-108))/6

(1±√-107)/6

1±(-10.34)/6

1+10.34/6 or 1-10.34/6

11.34/6 or -9.34/6

1.89 or 1.56