Algebra Examples

$x^{2} - 5 x + 6 = y$ , $(0, 3)$

Step 1

Rewrite the equation as $y = x^{2} - 5 x + 6$ .

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

Step 2

The Intermediate Value Theorem states that, if $f$ is a real-valued continuous function on the interval $[a, b]$ , and $u$ is a number between $f (a)$ and $f (b)$ , then there is a $c$ contained in the interval $[a, b]$ such that $f (c) = u$ .

$u = f (c) = 0$

Step 3

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$

Step 4

Calculate $f (a) = f (0) = {(0)}^{2} - 5 \cdot 0 + 6$ .

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Step 4.1

Simplify each term.

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Step 4.1.1

Raising $0$ to any positive power yields $0$ .

$f (0) = 0 - 5 \cdot 0 + 6$

Step 4.1.2

Multiply $- 5$ by $0$ .

$f (0) = 0 + 0 + 6$

Step 4.2

Simplify by adding numbers.

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Step 4.2.1

Add $0$ and $0$ .

$f (0) = 0 + 6$

Step 4.2.2

Add $0$ and $6$ .

$f (0) = 6$

Step 5

Calculate $f (b) = f (3) = {(3)}^{2} - 5 \cdot 3 + 6$ .

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Step 5.1

Simplify each term.

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Step 5.1.1

Raise $3$ to the power of $2$ .

$f (3) = 9 - 5 \cdot 3 + 6$

Step 5.1.2

Multiply $- 5$ by $3$ .

$f (3) = 9 - 15 + 6$

Step 5.2

Simplify by adding and subtracting.

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Step 5.2.1

Subtract $15$ from $9$ .

$f (3) = - 6 + 6$

Step 5.2.2

Add $- 6$ and $6$ .

$f (3) = 0$

Step 6

Since $0$ is on the interval $[0, 6]$ , solve the equation for $x$ at the root by setting $y$ to $0$ in $y = x^{2} - 5 x + 6$ .

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Step 6.1

Rewrite the equation as $x^{2} - 5 x + 6 = 0$ .

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

Step 6.2

Factor $x^{2} - 5 x + 6$ using the AC method.

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Step 6.2.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $6$ and whose sum is $- 5$ .

$- 3, - 2$

Step 6.2.2

Write the factored form using these integers.

$(x - 3) (x - 2) = 0$

Step 6.3

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 3 = 0$

$x - 2 = 0$

Step 6.4

Set $x - 3$ equal to $0$ and solve for $x$ .

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Step 6.4.1

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Step 6.4.2

Add $3$ to both sides of the equation.

$x = 3$

Step 6.5

Set $x - 2$ equal to $0$ and solve for $x$ .

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Step 6.5.1

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Step 6.5.2

Add $2$ to both sides of the equation.

$x = 2$

Step 6.6

The final solution is all the values that make $(x - 3) (x - 2) = 0$ true.

$x = 3, 2$

Step 7

The Intermediate Value Theorem states that there is a root $f (c) = 0$ on the interval $[0, 6]$ because $f$ is a continuous function on $[0, 3]$ .

The roots on the interval $[0, 3]$ are located at $x = 2$ .

Step 8