Algebra Examples

(1, 3)

$(1, 3)$

Step 1

x = 1

$x = 1$ and

x = 3

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

x - 1

$x - 1$ and

x - 3

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

(x - 1) (x - 3) = 0

$(x - 1) (x - 3) = 0$

Step 2

Expand

(x - 1) (x - 3)

$(x - 1) (x - 3)$ using the FOIL Method.

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Step 2.1

Apply the distributive property.

x (x - 3) - 1 (x - 3) = 0

$x (x - 3) - 1 (x - 3) = 0$

Step 2.2

Apply the distributive property.

x \cdot x + x \cdot - 3 - 1 (x - 3) = 0

$x \cdot x + x \cdot - 3 - 1 (x - 3) = 0$

Step 2.3

Apply the distributive property.

x \cdot x + x \cdot - 3 - 1 x - 1 \cdot - 3 = 0

$x \cdot x + x \cdot - 3 - 1 x - 1 \cdot - 3 = 0$

x \cdot x + x \cdot - 3 - 1 x - 1 \cdot - 3 = 0

$x \cdot x + x \cdot - 3 - 1 x - 1 \cdot - 3 = 0$

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 - 1 x - 1 \cdot - 3 = 0

$x^{2} + x \cdot - 3 - 1 x - 1 \cdot - 3 = 0$

Step 3.1.2

Move

- 3

$- 3$ to the left of

x

$x$ .

x^{2} - 3 \cdot x - 1 x - 1 \cdot - 3 = 0

$x^{2} - 3 \cdot x - 1 x - 1 \cdot - 3 = 0$

Step 3.1.3

Rewrite

- 1 x

$- 1 x$ as

- x

$- x$ .

x^{2} - 3 x - x - 1 \cdot - 3 = 0

$x^{2} - 3 x - x - 1 \cdot - 3 = 0$

Step 3.1.4

Multiply

- 1

$- 1$ by

- 3

$- 3$ .

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

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

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

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

Step 3.2

Subtract

x

$x$ from

- 3 x

$- 3 x$ .

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

$x^{2} - 4 x + 3 = 0$

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

$x^{2} - 4 x + 3 = 0$

Step 4

The standard quadratic equation using the given set of solutions

{1, 3}

${1, 3}$ is

y = x^{2} - 4 x + 3

$y = x^{2} - 4 x + 3$ .

y = x^{2} - 4 x + 3

$y = x^{2} - 4 x + 3$

Step 5

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