Precalculus Examples

$(2, - \frac{5 π}{6})$

Step 1

$x = 2$ and $x = - \frac{5 π}{6}$ are the two real distinct solutions for the quadratic equation, which means that $x - 2$ and $x + \frac{5 π}{6}$ are the factors of the quadratic equation.

$(x - 2) (x + \frac{5 π}{6}) = 0$

Step 2

Expand $(x - 2) (x + \frac{5 π}{6})$ using the FOIL Method.

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Step 2.1

Apply the distributive property.

$x (x + \frac{5 π}{6}) - 2 (x + \frac{5 π}{6}) = 0$

Step 2.2

Apply the distributive property.

$x \cdot x + x (\frac{5 π}{6}) - 2 (x + \frac{5 π}{6}) = 0$

Step 2.3

Apply the distributive property.

$x \cdot x + x (\frac{5 π}{6}) - 2 x - 2 \frac{5 π}{6} = 0$

Step 3

Simplify each term.

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Step 3.1

Multiply $x$ by $x$ .

$x^{2} + x (\frac{5 π}{6}) - 2 x - 2 \frac{5 π}{6} = 0$

Step 3.2

Combine $x$ and $\frac{5 π}{6}$ .

$x^{2} + \frac{x (5 π)}{6} - 2 x - 2 \frac{5 π}{6} = 0$

Step 3.3

Move $5$ to the left of $x$ .

$x^{2} + \frac{5 \cdot (x π)}{6} - 2 x - 2 \frac{5 π}{6} = 0$

Step 3.4

Cancel the common factor of $2$ .

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Step 3.4.1

Factor $2$ out of $- 2$ .

$x^{2} + \frac{5 x π}{6} - 2 x + 2 (- 1) (\frac{5 π}{6}) = 0$

Step 3.4.2

Factor $2$ out of $6$ .

$x^{2} + \frac{5 x π}{6} - 2 x + 2 \cdot (- 1 \frac{5 π}{2 \cdot 3}) = 0$

Step 3.4.3

Cancel the common factor.

$x^{2} + \frac{5 x π}{6} - 2 x + 2 \cdot (- 1 \frac{5 π}{2 \cdot 3}) = 0$

Step 3.4.4

Rewrite the expression.

$x^{2} + \frac{5 x π}{6} - 2 x - 1 \frac{5 π}{3} = 0$

Step 3.5

Rewrite $- 1 \frac{5 π}{3}$ as $- \frac{5 π}{3}$ .

$x^{2} + \frac{5 x π}{6} - 2 x - \frac{5 π}{3} = 0$

Step 4

The standard quadratic equation using the given set of solutions ${2, - \frac{5 π}{6}}$ is $y = x^{2} + \frac{5 x π}{6} - 2 x - \frac{5 π}{3}$ .

$y = x^{2} + \frac{5 x π}{6} - 2 x - \frac{5 π}{3}$

Step 5