Precalculus Examples

$(6, 4)$

$x = 6$ and $x = 4$ are the two real distinct solutions for the quadratic equation, which means that $x - 6$ and $x - 4$ are the factors of the quadratic equation.

$(x - 6) (x - 4) = 0$

Expand $(x - 6) (x - 4)$ using the FOIL Method.

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Apply the distributive property.

$x (x - 4) - 6 (x - 4) = 0$

Apply the distributive property.

$x x + x \cdot - 4 - 6 (x - 4) = 0$

Apply the distributive property.

$x x + x \cdot - 4 + (- 6 x - 6 \cdot - 4) = 0$

Remove parentheses.

$x x + x \cdot - 4 - 6 x - 6 \cdot - 4 = 0$

Simplify and combine like terms.

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Simplify each term.

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

$x^{1 + 1} + x \cdot - 4 - 6 x - 6 \cdot - 4 = 0$

Add $1$ and $1$ to get $2$ .

$x^{2} + x \cdot - 4 - 6 x - 6 \cdot - 4 = 0$

Move $- 4$ to the left of the expression $x \cdot - 4$ .

$x^{2} - 4 \cdot x - 6 x - 6 \cdot - 4 = 0$

Multiply $- 4$ by $x$ to get $- 4 x$ .

$x^{2} - 4 x - 6 x - 6 \cdot - 4 = 0$

Multiply $- 6$ by $- 4$ to get $24$ .

$x^{2} - 4 x - 6 x + 24 = 0$

Subtract $6 x$ from $- 4 x$ to get $- 10 x$ .

$x^{2} - 10 x + 24 = 0$

The standard quadratic equation using the given set of solutions ${6, 4}$ is $y = x^{2} - 10 x + 24$ .

$y = x^{2} - 10 x + 24$

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