Precalculus Examples

$(9, 7)$

$x = 9$ and $x = 7$ are the two real distinct solutions for the quadratic equation, which means that $x - 9$ and $x - 7$ are the factors of the quadratic equation.

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

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

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Apply the distributive property.

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

Apply the distributive property.

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

Apply the distributive property.

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

Remove parentheses.

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

Simplify and combine like terms.

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Simplify each term.

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Multiply $x$ by $x$ by adding the exponents.

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Combine $x$ and $x$

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Raise $x$ to the power of $1$ .

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

Raise $x$ to the power of $1$ .

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

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$x^{1 + 1} + x \cdot - 7 - 9 x - 9 \cdot - 7 = 0$

Add $1$ and $1$ .

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

Move $- 7$ to the left of the expression $x \cdot - 7$ .

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

Multiply $- 7$ by $x$ .

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

Multiply $- 9$ by $- 7$ .

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

Subtract $9 x$ from $- 7 x$ .

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

The standard quadratic equation using the given set of solutions ${9, 7}$ is $y = x^{2} - 16 x + 63$ .

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

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