Algebra Examples

(6, - 3)

$(6, - 3)$

Step 1

x = 6

$x = 6$ and

x = - 3

$x = - 3$ are the two real distinct solutions for the quadratic equation, which means that

x - 6

$x - 6$ and

x + 3

$x + 3$ are the factors of the quadratic equation.

(x - 6) (x + 3) = 0

$(x - 6) (x + 3) = 0$

Step 2

Expand

(x - 6) (x + 3)

$(x - 6) (x + 3)$ using the FOIL Method.

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Step 2.1

Apply the distributive property.

x (x + 3) - 6 (x + 3) = 0

$x (x + 3) - 6 (x + 3) = 0$

Step 2.2

Apply the distributive property.

x \cdot x + x \cdot 3 - 6 (x + 3) = 0

$x \cdot x + x \cdot 3 - 6 (x + 3) = 0$

Step 2.3

Apply the distributive property.

x \cdot x + x \cdot 3 - 6 x - 6 \cdot 3 = 0

$x \cdot x + x \cdot 3 - 6 x - 6 \cdot 3 = 0$

x \cdot x + x \cdot 3 - 6 x - 6 \cdot 3 = 0

$x \cdot x + x \cdot 3 - 6 x - 6 \cdot 3 = 0$

Step 3

Simplify and combine like terms.

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Step 3.1

Simplify each term.

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Step 3.1.1

Multiply

x

$x$ by

x

$x$ .

x^{2} + x \cdot 3 - 6 x - 6 \cdot 3 = 0

$x^{2} + x \cdot 3 - 6 x - 6 \cdot 3 = 0$

Step 3.1.2

Move

3

$3$ to the left of

x

$x$ .

x^{2} + 3 \cdot x - 6 x - 6 \cdot 3 = 0

$x^{2} + 3 \cdot x - 6 x - 6 \cdot 3 = 0$

Step 3.1.3

Multiply

- 6

$- 6$ by

3

$3$ .

x^{2} + 3 x - 6 x - 18 = 0

$x^{2} + 3 x - 6 x - 18 = 0$

x^{2} + 3 x - 6 x - 18 = 0

$x^{2} + 3 x - 6 x - 18 = 0$

Step 3.2

Subtract

6 x

$6 x$ from

3 x

$3 x$ .

x^{2} - 3 x - 18 = 0

$x^{2} - 3 x - 18 = 0$

x^{2} - 3 x - 18 = 0

$x^{2} - 3 x - 18 = 0$

Step 4

The standard quadratic equation using the given set of solutions

{6, - 3}

${6, - 3}$ is

y = x^{2} - 3 x - 18

$y = x^{2} - 3 x - 18$ .

y = x^{2} - 3 x - 18

$y = x^{2} - 3 x - 18$

Step 5

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