Problem 59
Question
Solve the equation or write no solution. Write the solutions as integers if possible. Otherwise write them as radical expressions. $$x^{2}=-9$$
Step-by-Step Solution
Verified Answer
The solutions to the equation \(x^{2}=-9\) are \(x = 3i\) and \(x = -3i\).
1Step 1: Understanding the equation
The equation given is \(x^{2}=-9\). In order to solve for \(x\), one must take the square root of both sides.
2Step 2: Applying square root
Taking the square root of \(x^{2}\) on the left side gives us \(x\). However, taking the square root of \(-9\) on the right side is not a real number. In this case, we must use the concept of imaginary numbers. Remember that \(\sqrt{-1}\) is represented by \(i\). Hence, \(\sqrt{-9} = \sqrt{9} \cdot \sqrt{-1} = 3i\).
3Step 3: Writing out the solutions
The solutions to the equation are \(\pm 3i\). This means \(x = 3i\) and \(x = -3i\).
Key Concepts
Complex NumbersImaginary NumbersRadical Expressions
Complex Numbers
The concept of complex numbers extends the idea of numbers beyond the real number line. A complex number is a combination of a real part and an imaginary part. It is expressed in the form \(a + bi\), where \(a\) and \(b\) are real numbers, and \(i\) is the imaginary unit.
In our exercise, the number \(-9\) required us to use complex numbers to find a solution. This results in two possible solutions, expressed as \(3i\) and \(-3i\). Complex numbers like these are crucial in many fields of science and engineering, as they provide a way to solve a wide range of problems.
- The real part \(a\) is any real number.
- The imaginary part \(bi\) involves the square root of a negative number.
In our exercise, the number \(-9\) required us to use complex numbers to find a solution. This results in two possible solutions, expressed as \(3i\) and \(-3i\). Complex numbers like these are crucial in many fields of science and engineering, as they provide a way to solve a wide range of problems.
Imaginary Numbers
Imaginary numbers arise when we take the square root of a negative number. Historically, the existence of such numbers was met with skepticism because they did not fit within the traditional real number framework.
When we defined \(i\) as the square root of \(-1\), mathematicians created a consistent way to extend the number system. This allows us to perform algebraic operations involving square roots of negative numbers, paving the way for a broader understanding of mathematics.
In the equation \(x^2 = -9\), we utilized \(i\) to solve it. Specifically, taking the square root of \(-9\) involves breaking it into \(\sqrt{9} \cdot \sqrt{-1}\), which translates to \(3i\). Imaginary numbers transform problems where no real solutions exist, creating complex but feasible solutions.
When we defined \(i\) as the square root of \(-1\), mathematicians created a consistent way to extend the number system. This allows us to perform algebraic operations involving square roots of negative numbers, paving the way for a broader understanding of mathematics.
In the equation \(x^2 = -9\), we utilized \(i\) to solve it. Specifically, taking the square root of \(-9\) involves breaking it into \(\sqrt{9} \cdot \sqrt{-1}\), which translates to \(3i\). Imaginary numbers transform problems where no real solutions exist, creating complex but feasible solutions.
Radical Expressions
Radical expressions involve roots of numbers, where the square root is one of the most common types. A radical expression like \(\sqrt{a}\) asks what number squared will result in \(a\).
Radical expressions extend to both positive and negative numbers, but with negative numbers, they generate imaginary numbers as seen in the equation \(x^2 = -9\). By expressing \(\sqrt{-9}\) as \(\sqrt{9} \cdot \sqrt{-1}\), we handle the negative sign separately by introducing \(i\).
When simplifying such an expression, especially in quadratic equations, it often leads to a complex number. Hence, solutions to these equations might not always be integers but are neatly expressed as radical expressions, embracing the inclusion of complex elements. This approach gives us a complete set of solutions to otherwise insolvable problems involving negative roots.
Radical expressions extend to both positive and negative numbers, but with negative numbers, they generate imaginary numbers as seen in the equation \(x^2 = -9\). By expressing \(\sqrt{-9}\) as \(\sqrt{9} \cdot \sqrt{-1}\), we handle the negative sign separately by introducing \(i\).
When simplifying such an expression, especially in quadratic equations, it often leads to a complex number. Hence, solutions to these equations might not always be integers but are neatly expressed as radical expressions, embracing the inclusion of complex elements. This approach gives us a complete set of solutions to otherwise insolvable problems involving negative roots.
Other exercises in this chapter
Problem 58
Which is the simplified form of \(4 \frac{\sqrt{125}}{\sqrt{25}} ?\) (A) \(2 \sqrt{5}\) (B) \(4 \sqrt{5}\) (C) \(20 \sqrt{5}\) (D) \(\frac{4 \sqrt{5}}{5}\)
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SKETCHING GRAPHS Sketch the graph of the function. Label the vertex. $$ y=-4 x^{2}+32 x-20 $$
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The solution of a quadratic equation can be found by graphing each side separately and locating the points of intersection. You may wish to consult page 532 for
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Find the x-intercepts of the graph of the equation. $$y=-3 x^{2}-2 x+1$$
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