Precalculus Examples

(9, 7)

$(9, 7)$

Step 1

x = 9

$x = 9$ and

x = 7

$x = 7$ are the two real distinct solutions for the quadratic equation, which means that

x - 9

$x - 9$ and

x - 7

$x - 7$ are the factors of the quadratic equation.

(x - 9) (x - 7) = 0

$(x - 9) (x - 7) = 0$

Step 2

Expand

(x - 9) (x - 7)

$(x - 9) (x - 7)$ using the FOIL Method.

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Step 2.1

Apply the distributive property.

x (x - 7) - 9 (x - 7) = 0

$x (x - 7) - 9 (x - 7) = 0$

Step 2.2

Apply the distributive property.

x \cdot x + x \cdot - 7 - 9 (x - 7) = 0

$x \cdot x + x \cdot - 7 - 9 (x - 7) = 0$

Step 2.3

Apply the distributive property.

x \cdot x + x \cdot - 7 - 9 x - 9 \cdot - 7 = 0

$x \cdot x + x \cdot - 7 - 9 x - 9 \cdot - 7 = 0$

x \cdot x + x \cdot - 7 - 9 x - 9 \cdot - 7 = 0

$x \cdot x + x \cdot - 7 - 9 x - 9 \cdot - 7 = 0$

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 - 7 - 9 x - 9 \cdot - 7 = 0

$x^{2} + x \cdot - 7 - 9 x - 9 \cdot - 7 = 0$

Step 3.1.2

Move

- 7

$- 7$ to the left of

x

$x$ .

x^{2} - 7 \cdot x - 9 x - 9 \cdot - 7 = 0

$x^{2} - 7 \cdot x - 9 x - 9 \cdot - 7 = 0$

Step 3.1.3

Multiply

- 9

$- 9$ by

- 7

$- 7$ .

x^{2} - 7 x - 9 x + 63 = 0

$x^{2} - 7 x - 9 x + 63 = 0$

x^{2} - 7 x - 9 x + 63 = 0

$x^{2} - 7 x - 9 x + 63 = 0$

Step 3.2

Subtract

9 x

$9 x$ from

- 7 x

$- 7 x$ .

x^{2} - 16 x + 63 = 0

$x^{2} - 16 x + 63 = 0$

x^{2} - 16 x + 63 = 0

$x^{2} - 16 x + 63 = 0$

Step 4

The standard quadratic equation using the given set of solutions

{9, 7}

${9, 7}$ is

y = x^{2} - 16 x + 63

$y = x^{2} - 16 x + 63$ .

y = x^{2} - 16 x + 63

$y = x^{2} - 16 x + 63$

Step 5

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