Chapter 6: Problem 32
Use the method of completing the square to solve each quadratic equation. $$y^{2}-9 y+30=0$$
Short Answer
Expert verified
The solutions are \( y = \frac{9}{2} \pm \frac{\sqrt{39}i}{2} \).
Step by step solution
01
Identify the Quadratic Equation
The given quadratic equation is \( y^2 - 9y + 30 = 0 \). Notice that it is in the form \( ay^2 + by + c = 0 \) where \( a = 1 \), \( b = -9 \), and \( c = 30 \).
02
Move Constant Term to the Right Side
Start by moving the constant term to the right side of the equation: \( y^2 - 9y = -30 \). This will prepare the equation for completing the square.
03
Find the Value to Complete the Square
To complete the square, take half of the coefficient of \( y \) (which is -9), square it, and add it to both sides. Half of -9 is \( \frac{-9}{2} = -\frac{9}{2} \). Squaring it, we get \( \left(-\frac{9}{2}\right)^2 = \frac{81}{4} \).
04
Add and Subtract the Square Value
Add and subtract \( \frac{81}{4} \) inside the equation: \( y^2 - 9y + \frac{81}{4} = -30 + \frac{81}{4} \). Now the left side becomes a perfect square trinomial.
05
Rewrite as a Perfect Square
Rewrite the left side as \( (y - \frac{9}{2})^2 \): \( (y - \frac{9}{2})^2 = -30 + \frac{81}{4} \).
06
Simplify the Right Side
Combine the terms on the right: Convert -30 to a fraction with a denominator of 4: \( -30 = \frac{-120}{4} \). Therefore, \( \frac{-120}{4} + \frac{81}{4} = \frac{-39}{4} \).
07
Solve for y by Taking the Square Root
Take the square root of both sides: \( y - \frac{9}{2} = \pm \sqrt{\frac{-39}{4}} \). Since the square root of a negative number involves \( i \), we have \( y - \frac{9}{2} = \pm \frac{\sqrt{39}i}{2} \).
08
Solve for y
Isolate \( y \) by adding \( \frac{9}{2} \) to both sides: \( y = \frac{9}{2} \pm \frac{\sqrt{39}i}{2} \).
09
Write the Final Solution
The solutions are \( y = \frac{9}{2} + \frac{\sqrt{39}i}{2} \) and \( y = \frac{9}{2} - \frac{\sqrt{39}i}{2} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Equation
A quadratic equation is a mathematical expression of the form \( ax^2 + bx + c = 0 \) where \( a, b, \) and \( c \) are constants, and \( a eq 0 \). This equation represents a parabola on a graph. In our exercise, the equation presented was \( y^2 - 9y + 30 = 0 \). Here, we see:
- \( a = 1 \), indicating that the parabola opens upwards.
- \( b = -9 \), which affects the axis of symmetry and vertex position.
- \( c = 30 \), the y-intercept of the quadratic graph.
Perfect Square Trinomial
A perfect square trinomial is a special form of quadratic expressions. It is written as \((x + d)^2\) or \((x - d)^2\), and when expanded, it equals \(x^2 + 2dx + d^2\). In our scenario, we transformed \(y^2 - 9y\) into a perfect square trinomial.
To do this, we:
To do this, we:
- Identified the coefficient of \(y\), which was \(-9\).
- Calculated half of \(-9\), yielding \(-\frac{9}{2}\).
- Squared \(-\frac{9}{2}\) to get \(\frac{81}{4}\).
Imaginary Numbers
Imaginary numbers arise when we take the square root of a negative number. In our solved equation, taking the square root of both sides resulted in the expression \( \pm \sqrt{\frac{-39}{4}} \).Since square roots of negative numbers are not defined within the realm of real numbers, we use the imaginary unit \( i \), where \( i = \sqrt{-1} \). This allows us to write the expression as \( \pm \frac{\sqrt{39}i}{2} \).
In conclusion, when solving \((y - \frac{9}{2})^2 = -\frac{39}{4}\), you end up with complex solutions due to this imaginary number. The final solutions are expressed as \( y = \frac{9}{2} \pm \frac{\sqrt{39}i}{2} \).
Understanding imaginary numbers is crucial for solving equations that don't intersect the x-axis, as these imply that our solutions exist as complex numbers with both real and imaginary components.
In conclusion, when solving \((y - \frac{9}{2})^2 = -\frac{39}{4}\), you end up with complex solutions due to this imaginary number. The final solutions are expressed as \( y = \frac{9}{2} \pm \frac{\sqrt{39}i}{2} \).
Understanding imaginary numbers is crucial for solving equations that don't intersect the x-axis, as these imply that our solutions exist as complex numbers with both real and imaginary components.