Using the Quadratic Formula to find Complex Roots

We will solve the given quadratic equation by using the Quadratic Formula.

a x^{2} + b x + c = 0 \Leftrightarrow x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

Let's identify the values of

a,

b,

and

c

in our equation.

3 x^{2} + (- 2) x + (- 4) = 0

We can see above that

a = 3,

b = - 2,

and

c = - 4 .

Now we can substitute these values into the Quadratic Formula.

x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

SubstituteValuesSubstitute values

x = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 ( 3 ) ( - 4 )}{2 ( 3 )}

Simplify

- (- a) = a

x = \frac{2 \pm ( - 2 ) ^{2} - 4 ( 3 ) ( - 4 )}{2 ( 3 )}

CalcPowCalculate power

x = \frac{2 \pm 4 - 4 ( 3 ) ( - 4 )}{2 ( 3 )}

(- a) b = - a b

x = \frac{2 \pm 4 - 1 2 ( - 4 )}{2 ( 3 )}

- a (- b) = a \cdot b

x = \frac{2 \pm 4 + 4 8}{2 ( 3 )}

SubTermSubtract term

x = \frac{2 \pm 5 2}{2 ( 3 )}

x = \frac{2 \pm 5 2}{6}

Calculate root

SplitIntoFactorsSplit into factors

x = \frac{2 \pm 4 \cdot 1 3}{6}

a \cdot b = a \cdot b

x = \frac{2 \pm 4 \cdot 1 3}{6}

CalcRootCalculate root

x = \frac{2 \pm 2 1 3}{6}

Simplify

FactorOutFactor out

2

x = \frac{2 ( 1 \pm 1 3 )}{6}

CancelCommonFacCancel out common factors

x = \frac{1 \pm 1 3}{3}

The equation has two solutions and they are

x = \frac{1 + 1 3}{3}

and

x = \frac{1 - 1 3}{3} .

Using the Quadratic Formula to find Complex Roots 1.5 - Solution