Explain each step of solving the problem x² + x - 6 = 0 along with illustrations (if available).
Understand the Problem
The question is asking to solve the quadratic equation x² + x - 6 = 0 and to provide an explanation of each step along with possible illustrations to enhance understanding.
Answer
The solutions to the equation are $x = 2$ and $x = -3$.
Answer for screen readers
The solutions to the equation $x^2 + x - 6 = 0$ are $x = 2$ and $x = -3$.
Steps to Solve
- Identify the standard form of a quadratic equation
The equation $x^2 + x - 6 = 0$ is already in the standard form of a quadratic equation, which is $ax^2 + bx + c = 0$. Here, $a = 1$, $b = 1$, and $c = -6$.
- Determine the values of a, b, and c
From the equation, we identify:
- $a = 1$
- $b = 1$
- $c = -6$
- Apply the quadratic formula
To find the roots of the equation, we use the quadratic formula: $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$
- Calculate the discriminant
First, we calculate the discriminant $b^2 - 4ac$: $$ b^2 - 4ac = 1^2 - 4(1)(-6) = 1 + 24 = 25 $$
- Substitute values into the quadratic formula
Now we substitute $a$, $b$, and the discriminant into the quadratic formula: $$ x = \frac{-1 \pm \sqrt{25}}{2(1)} $$
- Calculate the roots
Next, we find the value of $\sqrt{25}$, which is $5$: $$ x = \frac{-1 \pm 5}{2} $$
- Find the two possible solutions
Now we calculate the two possible solutions:
- For $x_1 = \frac{-1 + 5}{2} = \frac{4}{2} = 2$
- For $x_2 = \frac{-1 - 5}{2} = \frac{-6}{2} = -3$
The solutions to the equation $x^2 + x - 6 = 0$ are $x = 2$ and $x = -3$.
More Information
The roots of a quadratic equation represent the x-values where the function intersects the x-axis. In this case, the graph of the equation $y = x^2 + x - 6$ will cross the x-axis at $(2, 0)$ and $(-3, 0)$. Quadratic equations can have either two real roots, one real root, or no real roots depending on the value of the discriminant.
Tips
- Miscalculating the discriminant: Always ensure to double-check your calculations for the discriminant $b^2 - 4ac$.
- Forgetting to consider both the $+$ and $-$ in the quadratic formula, which results in missing one of the roots.
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