Solving Equations

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break it up

Common Factor

Solve $\sqrt {x+9} + x \cdot \frac {1}{2} (x+9)^{-\tfrac {1}{2}} = 0$

explanation

A common factor is a factor of the least degree in all of the terms. We have one here. $(x+9)$ is a factor in both terms. The powers are $\frac {1}{2}$ and $-\frac {1}{2}$. The least of which is $-\frac {1}{2}$.

We will use the distributive property to factor out $(x+9)^{\answer {-\frac {1}{2}}}$.

$(x+9)^{-\tfrac {1}{2}} \left ((x+9) + x \cdot \frac {1}{2}\right ) = 0$

We could, in turn, write this as a fraction

\[ \frac {\left ((x+9) + x \cdot \frac {1}{2}\right )}{\sqrt {x+9}} = 0 \]

\[ \frac { \frac {3}{2} x + 9}{\sqrt {x+9}} = 0 \]

This is a fraction, so it equals $0$ when the numerator equals $0$ and the denominator does not equal $0$.

$\answer {\frac {3}{2} x + 9} = 0$

$x = -6$, which is in the domain of $\sqrt {x+9} + x \cdot \frac {1}{2} (x+9)^{-\tfrac {1}{2}}$.

Or,

Or, we could have changed the negative exponents into fractions from the beginning.

\[ \sqrt {x+9} + x \cdot \frac {1}{2} (x+9)^{-\tfrac {1}{2}} = \sqrt {x+9} + \frac {x}{2 \sqrt {x+9}} = \frac {(x+9)+x}{2 \sqrt {x+9}} = \frac {2x+9}{2 \sqrt {x+9}} \]

A fraction equals $0$ when the numerator equals $0$.

Incidentally, $\sqrt {x+9} + x \cdot \frac {1}{2} (x+9)^{-\tfrac {1}{2}}$ is the derivative of $\sqrt {x+9}$.

So, we are finding where $\sqrt {x+9}$ switches between increasing and decreasing, i.e. critical numbers.

Common Denominator

explanation

Factoring out a common factor can also be viewed as getting a common denominator and combining into a single fraction.

\[ \sqrt {x+9} + x \cdot \frac {1}{2} (x+9)^{-\tfrac {1}{2}} = \sqrt {x+9} + x \cdot \frac {1}{2 \, \sqrt {x+9}} = \frac {2(x+9)}{2 \, \sqrt {x+9}} + \frac {x}{2 \, \sqrt {x+9}} \]

\[ = \frac {2 \, (x + 9) + x}{2 \, \sqrt {x+9}} \]

\[ = \frac {3x + 18}{2 \, \sqrt {x+9}} \]

\[ = \frac {3 \, (x + 6)}{2 \, \sqrt {x+9}} \]

Common Factor

Solve

\[ \frac {x^3 (18x) - 9(x^2-3)(3x^2)}{(x^3)^2} = 0 \]

explanation

The numerator is a difference of two terms and $9 x^2$ is a common factor. We can use the Distributive Property to factor it out.

\[ \frac {9x^2 (2x^2 - 3(x^2-3))}{x^6} = 0 \]

\[ \frac {9 \left ( \answer {9 - x^2} \right )}{x^4} = 0 \]

This is a fraction, so it equals $0$ when the numerator equals $0$ and the denominator does not equal $0$.

$9 - x^2 = 0$

This has two solutions: $\{ -3, 3 \}$, both of which are in the domain.

Incidentally, $\frac {x^3 (18x) - 9(x^2-3)(3x^2)}{(x^3)^2}$ is the derivative of $\frac {9(x^2-3)}{x^3}$.

So, we are finding where $\frac {9(x^2-3)}{x^3}$ switches between increasing and decreasing, i.e. critical numbers.

Common Factor

Solve $12 x^3 - 12 x^2 = 0 $

explanation

The Distributive Property (factoring) gives us the product $12 x^2 (x-1)$.

We have two solutions: $\{ 0, 1 \}$, both of which are in the domain.

$12 x^3 - 12 x^2$ is the derivative of $3x^4 - 4x^3$. Therefore, the graph of $y = 3x^4 - 4x^3$ has horizontal tangent lines at $(0, 0)$ and $(1, -1)$.

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Fractions

Solve $2 - \frac {2}{\sqrt [3]{x}} = 0$

explanation

Converting to exponents gives us $2 - \frac {2}{x^{\tfrac {1}{3}}} = 0$

$2 = \frac {2}{x^{\tfrac {1}{3}}}$

The only way this can happen is if $x^{\tfrac {1}{3}} = \answer {1}$.

This has one solution: $\{ 1 \}$

Or,

Or, we could have combined terms from the beginning into a fraction.

\[ 2 - \frac {2}{\sqrt [3]{x}} = \frac {4\sqrt [3]{x}}{\sqrt [3]{x}} - \frac {2}{\sqrt [3]{x}} = \frac {4\sqrt [3]{x}-2}{\sqrt [3]{x}} \]

A fraction equals $0$ when the numerator equals $0$.

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more examples can be found by following this link
More Examples of Function Zeros

2026-05-31 19:29:08