Examples

(1, 5)

$(1, 5)$

Step 1

x = 1

$x = 1$ and

x = 5

$x = 5$ are the two real distinct solutions for the quadratic equation, which means that

x - 1

$x - 1$ and

x - 5

$x - 5$ are the factors of the quadratic equation.

(x - 1) (x - 5) = 0

$(x - 1) (x - 5) = 0$

Step 2

Expand

(x - 1) (x - 5)

$(x - 1) (x - 5)$ using the FOIL Method.

Tap for more steps...

Step 2.1

Apply the distributive property.

x (x - 5) - 1 (x - 5) = 0

$x (x - 5) - 1 (x - 5) = 0$

Step 2.2

Apply the distributive property.

x \cdot x + x \cdot - 5 - 1 (x - 5) = 0

$x \cdot x + x \cdot - 5 - 1 (x - 5) = 0$

Step 2.3

Apply the distributive property.

x \cdot x + x \cdot - 5 - 1 x - 1 \cdot - 5 = 0

$x \cdot x + x \cdot - 5 - 1 x - 1 \cdot - 5 = 0$

x \cdot x + x \cdot - 5 - 1 x - 1 \cdot - 5 = 0

$x \cdot x + x \cdot - 5 - 1 x - 1 \cdot - 5 = 0$

Step 3

Simplify and combine like terms.

Tap for more steps...

Step 3.1

Simplify each term.

Tap for more steps...

Step 3.1.1

Multiply

x

$x$ by

x

$x$ .

x^{2} + x \cdot - 5 - 1 x - 1 \cdot - 5 = 0

$x^{2} + x \cdot - 5 - 1 x - 1 \cdot - 5 = 0$

Step 3.1.2

Move

- 5

$- 5$ to the left of

x

$x$ .

x^{2} - 5 \cdot x - 1 x - 1 \cdot - 5 = 0

$x^{2} - 5 \cdot x - 1 x - 1 \cdot - 5 = 0$

Step 3.1.3

Rewrite

- 1 x

$- 1 x$ as

- x

$- x$ .

x^{2} - 5 x - x - 1 \cdot - 5 = 0

$x^{2} - 5 x - x - 1 \cdot - 5 = 0$

Step 3.1.4

Multiply

- 1

$- 1$ by

- 5

$- 5$ .

x^{2} - 5 x - x + 5 = 0

$x^{2} - 5 x - x + 5 = 0$

x^{2} - 5 x - x + 5 = 0

$x^{2} - 5 x - x + 5 = 0$

Step 3.2

Subtract

$x$ from

$- 5 x$ .

$x^{2} - 6 x + 5 = 0$

Step 4

The standard quadratic equation using the given set of solutions

${1, 5}$ is

$y = x^{2} - 6 x + 5$ .

$y = x^{2} - 6 x + 5$

Step 5

Enter YOUR Problem