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Nathaniel is using the quadratic formula to solve the given equation.

\[ x^2 \space + \space 5x \space – \space 6 \space = \space 0 $- $ X \space = \space \frac{-b+ \sqrt(b^2 – 4ac)}{2a} \space where \space a \space = \space -1,  \space b \space = \space 5  \space and  \space c \space = \space -6 \]

-What are the possible solutions to the given equation?

The main objective of this question is to find the solution to the given equation which is solved with the help of a quadratic equation.

This question uses the concept of a solution to the given equation. The collection of all values that, when used to replace unknowns, results in an accurate equation is known as the solution.

Expert Answer

The given equation is:

\[ x^2 \space + \space 5x \space – \space 6 \space = \space 0 \]

We know that:

\[X \space = \space \frac{-b \pm \sqrt(b^2 – 4ac)}{2a} where \space a \space = \space -1, \space b \space = \space 5 \space and \space c \space = \space -6 \]

By putting the values, we get:

\[X \space = \space \frac{-5 \pm \sqrt(25 – 4 ( 1 ) ( -6 )}{2 (1) }\]

\[X \space = \space \frac{-5 \pm \sqrt(25 + 24}{2 (1) }\]

\[X \space = \space \frac{-5 \pm \sqrt(25 + 24}{2 }\]

\[X \space = \space \frac{-5 \pm \sqrt(49}{2 }\]

Taking the square-root results in:

\[X \space = \space \frac{-5 \pm 7}{2 }\]

\[X \space = \space \frac{- 5  + 7}{2 }\]

\[X \space = \space \frac{- 5  – 7}{2 }\]

\[X \space = \space \frac{2}{2 } X\]

\[X \space = \space 1  \space and \space – 5 \]

Thus, the final answer is $ X \space = \space 1 $ and $ X \space = \space -5$.

Numerical Answer

The solution to the given equation which is solved with the quadratic formula is $ X \space = \space 1 $ & $ X \space = \space -5$.

Example

Find the solution to the given equation and solve it using the quadratic formula.

\[x^3 \space + \space 5x \space – \space 6 \space = \space 0]

The given equation is:

\[ x^3 \space + \space 5x \space – \space 6 \space = \space 0 \]

We know that:

\[X \space = \space \frac{-b \pm \sqrt(b^2 – 4ac)}{2a} where \space a \space = \space -1, \space b \space = \space 5 \space and \space c \space = \space -6 \]

By putting the values, we get:

\[X \space = \space \frac{-5 \pm \sqrt(25 – 4 ( 1 ) ( -6 )}{2 (1) }\]

\[X \space = \space \frac{-5 \pm \sqrt(25 + 24}{2 (1) }\]

\[X \space = \space \frac{-5 \pm \sqrt(25 + 24}{2 }\]

\[X \space = \space \frac{-5 \pm \sqrt(49}{2 }\]

Taking the square-root results in:

\[X \space = \space \frac{-5 \pm 7}{2 }\]

\[X \space = \space \frac{- 5  + 7}{2 }\]

\[X \space = \space \frac{- 5  – 7}{2 }\]

\[X \space = \space \frac{2}{2 } X\]

\[X \space = \space 1 \space  and \space – 5 \]

Thus, the final answer to the equation $ x^3 \space + \space 5x \space – \space 6 \space = \space 0 $is $ X \space = \space 1 $ & $ X \space = \space -5$.

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