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Using the Quadratic Formula to find Complex Roots

Quadratic Functions and Equations

Using the Quadratic Formula to find Complex Roots 1.9 - Solution

Return to Using the Quadratic Formula to find Complex Roots

a

Let's use the quadratic formula to find the solutions.

x^{2} - 6 x + 13 = 0

UseQuadFormUse the Quadratic Formula:

a = 1, b = - 6, c = 13

x = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 1 \cdot 1 3}{2 \cdot 1}

- (- a) = a

x = \frac{6 \pm ( - 6 ) ^{2} - 4 \cdot 1 \cdot 1 3}{2 \cdot 1}

CalcPowProdCalculate power and product

x = \frac{6 \pm 3 6 - 5 2}{2}

SubTermSubtract term

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

We have a negative number under the root sign. Therefore, the solutions will be complex.

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

- a = a \cdot i

x = \frac{6 \pm 1 6 \cdot i}{2}

CalcRootCalculate root

x = \frac{6 \pm 4 \cdot i}{2}

SimpQuotSimplify quotient

x = 3 \pm 2 i

The solutions are

x = 3 + 2 i

and

x = 3 - 2 i .

b

Now multiply the solutions together, which gives

(3 + 2 i) (3 - 2 i) .

This is a multiplication of two parentheses, with the only difference between them being a negative sign. That means it is a conjugate pair of binomials. Let's calculate its product.

(3 + 2 i) (3 - 2 i)

ExpandDiffSquares

(a + b) (a - b) = a^{2} - b^{2}

3^{2} - (2 i)^{2}

CalcPowCalculate power

9 - (2 i)^{2}

(a b)^{m} = a^{m} b^{m}

9 - 4 i^{2}

i^{2} = - 1

9 - (- 4)

a - (- b) = a + b

13

The product of the solutions is

13 .