Precalculus Examples

$(1, 3)$

$x = 1$ and $x = 3$ are the two real distinct solutions for the quadratic equation, which means that $x - 1$ and $x - 3$ are the factors of the quadratic equation.

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

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

Tap for more steps...

Apply the distributive property.

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

Apply the distributive property.

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

Apply the distributive property.

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

Remove parentheses.

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

Simplify and combine like terms.

Tap for more steps...

Simplify each term.

Tap for more steps...

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

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

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

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

Move $- 3$ to the left of the expression $x \cdot - 3$ .

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

Multiply $- 3$ by $x$ to get $- 3 x$ .

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

Rewrite $- 1 x$ as $- x$ .

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

Multiply $- 1$ by $- 3$ to get $3$ .

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

Subtract $x$ from $- 3 x$ to get $- 4 x$ .

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

The standard quadratic equation using the given set of solutions ${1, 3}$ is $y = x^{2} - 4 x + 3$ .

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

Enter YOUR Problem