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Magazine Chapter 9: Problem 47
Solve by extracting the roots and then solve by using the quadratic formula. Check answer. $$ x_{2}+9=0 $$
Short Answer
Expert verified
The solutions are \( x = 3i \) and \( x = -3i \).
Step by step solution
01
Isolate the Quadratic Term
First, we need to isolate the quadratic term in the given equation. The original equation is \[ x^2 + 9 = 0 \]Subtract 9 from both sides to get:\[ x^2 = -9 \]
02
Extract the Square Roots
With the quadratic term isolated, we extract the square roots of both sides to solve for \( x \).\[ x = \pm \sqrt{-9} \]Since the square root of \(-9\) involves imaginary numbers, we can express \( x \) as:\[ x = \pm 3i \]where \( i \) is the imaginary unit, defined by \( i^2 = -1 \).
03
Identify Coefficients for the Quadratic Formula
Next, we will verify the solution by using the quadratic formula. Start by identifying the coefficients of the general form \( ax^2 + bx + c = 0 \).For \( x^2 + 9 = 0 \), we have:\[ a = 1, \quad b = 0, \quad c = 9 \]
04
Apply the Quadratic Formula
The quadratic formula is given by:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Plugging in the coefficients from Step 3, we get:\[ x = \frac{-0 \pm \sqrt{0 - 4 \times 1 \times 9}}{2 \times 1} \]Simplify under the square root:\[ x = \pm \sqrt{-36} / 2 \]Which evaluates to:\[ x = \pm 3i \]
05
Verify the Solutions
Both methods of solving resulted in \( x = \pm 3i \). Hence the solutions are consistent and correct.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
quadratic formula
The quadratic formula is a reliable tool to find the solutions of a quadratic equation. It is applicable to any quadratic equation of the form \( ax^2 + bx + c = 0 \). The formula itself is expressed as:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This equation helps us find the values of \(x\) that satisfy the quadratic equation. Let's break it down:
In the problem \(x^2 + 9 = 0\), we determined \(a\), \(b\), and \(c\) as 1, 0, and 9, respectively. Substituting these into the quadratic formula confirms the presence of imaginary solutions.
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This equation helps us find the values of \(x\) that satisfy the quadratic equation. Let's break it down:
- \(a\), \(b\), and \(c\) are coefficients from the quadratic equation.
- \(-b\) changes the sign of \(b\), which is crucial for finding solutions.
- \(\pm\) indicates that there are generally two potential solutions: one with a plus and the other with a minus.
- The term under the square root sign, \(b^2 - 4ac\), is known as the discriminant. It tells us about the nature of roots:
- If the discriminant is positive, the roots are real and distinct.
- If it's zero, there's exactly one real root, called a "double root".
- If it's negative, the equation has complex roots involving imaginary numbers.
In the problem \(x^2 + 9 = 0\), we determined \(a\), \(b\), and \(c\) as 1, 0, and 9, respectively. Substituting these into the quadratic formula confirms the presence of imaginary solutions.
extracting square roots
Extracting square roots is a straightforward process for solving quadratic equations. This method specifically applies when the equation can be expressed in the form \(x^2 = k\). Here's how it works:
Now, take the square root of both sides, hence:\[x = \pm \sqrt{-9}\]This is where imaginary numbers come into play, as \(\sqrt{-9}\) is simplified to \(3i\). Thus we find \(x = \pm 3i\), indicating that the solutions are not real numbers, but indeed imaginary.
- First, isolate \(x^2\) on one side of the equation.
- The next step is to apply the square root to both sides.
- Remember, taking the square root introduces a \(\pm\) sign, leading to two possible solutions for \(x\).
Now, take the square root of both sides, hence:\[x = \pm \sqrt{-9}\]This is where imaginary numbers come into play, as \(\sqrt{-9}\) is simplified to \(3i\). Thus we find \(x = \pm 3i\), indicating that the solutions are not real numbers, but indeed imaginary.
quadratic equations
Quadratic equations are polynomial equations of the second degree, characterized by the general form \(ax^2 + bx + c = 0\). They are significant in various fields of science and engineering due to their occurrence in real-world situations. Let鈥檚 explore their core attributes:
In solving any quadratic equation, understanding these aspects helps in predicting the behavior and nature of its solutions. For \(x^2 + 9 = 0\), the quadratic opens upwards and does not intersect the x-axis, hence the imaginary roots \(x = \pm 3i\). This reflects a graph that does not cross or touch the x-axis, consistent with purely imaginary solutions.
- They always graph as parabolas, which can open upward or downward depending on the sign of \(a\). A positive \(a\) opens upwards, forming a U-shape, whereas a negative \(a\) opens downwards.
- The solutions to a quadratic equation, or roots, are the values of \(x\) where the parabola intersects the x-axis. These points can be real or complex numbers.
- The number of roots is determined by the discriminant \(b^2 - 4ac\). As previously mentioned, this tells us about the nature and amount of solutions a quadratic equation can have.
In solving any quadratic equation, understanding these aspects helps in predicting the behavior and nature of its solutions. For \(x^2 + 9 = 0\), the quadratic opens upwards and does not intersect the x-axis, hence the imaginary roots \(x = \pm 3i\). This reflects a graph that does not cross or touch the x-axis, consistent with purely imaginary solutions.
