Problem 39
Question
Solve the quadratic equation by using the quadratic formula. If the solutions are not real, state No Real Solution. $$ x^{2}+x=4 $$
Step-by-Step Solution
Verified Answer
The solutions are \(x = \frac{-1 + \sqrt{17}}{2}\) and \(x = \frac{-1 - \sqrt{17}}{2}\).
1Step 1: Rearrange the Equation
The quadratic equation must be in standard form, which is \(ax^2 + bx + c = 0\). Start by rearranging the given equation:\[ x^2 + x - 4 = 0 \]
2Step 2: Identify Coefficients
Identify the coefficients from the standard form of the equation \(ax^2 + bx + c = 0\). Here, \(a = 1\), \(b = 1\), and \(c = -4\).
3Step 3: Use the Quadratic Formula
The quadratic formula is given by:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Substitute \(a = 1\), \(b = 1\), and \(c = -4\) into the formula.
4Step 4: Calculate the Discriminant
The discriminant, \(b^2 - 4ac\), determines the nature of the roots. Calculate:\[ 1^2 - 4(1)(-4) = 1 + 16 = 17 \]Since 17 is positive, there are two distinct real solutions.
5Step 5: Solve for \(x\)
Substitute the values back into the quadratic formula:\[ x = \frac{-1 \pm \sqrt{17}}{2(1)} \]This simplifies to:\[ x = \frac{-1 + \sqrt{17}}{2} \quad \text{and} \quad x = \frac{-1 - \sqrt{17}}{2} \]
6Step 6: Present the Solutions
Thus, the solutions to the quadratic equation are:\[ x = \frac{-1 + \sqrt{17}}{2} \] and \[ x = \frac{-1 - \sqrt{17}}{2} \]
Key Concepts
quadratic formulareal solutionsdiscriminantstandard form of a quadratic equation
quadratic formula
The quadratic formula is a powerful tool to solve quadratic equations. It is especially useful when factoring the equation is complex or impossible. The formula is:
When using the quadratic formula, you substitute these coefficients into the formula to solve for \( x \). The \( \pm \) symbol indicates that the formula will give two possible solutions when the discriminant is positive, which means you will have two roots for the equation. Try to memorize the formula, as it is extremely helpful for solving a wide variety of quadratic equations.
- \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
When using the quadratic formula, you substitute these coefficients into the formula to solve for \( x \). The \( \pm \) symbol indicates that the formula will give two possible solutions when the discriminant is positive, which means you will have two roots for the equation. Try to memorize the formula, as it is extremely helpful for solving a wide variety of quadratic equations.
real solutions
Real solutions refer to the outputs of an equation that are real numbers, which are numbers that do not involve imaginary components (such as the square root of a negative number). When solving quadratic equations using the quadratic formula, the nature of the solutions—whether they are real or complex—depends largely on the value of the discriminant (the part under the square root in the formula, \( b^2 - 4ac \)).
If the discriminant is positive, there are two distinct real solutions. If it is zero, there is exactly one real solution or a repeated root. Conversely, if the discriminant is negative, no real solutions exist as the roots are complex numbers (involving imaginary numbers).
In our example, the discriminant was positive (17), meaning we have two distinct real solutions.
If the discriminant is positive, there are two distinct real solutions. If it is zero, there is exactly one real solution or a repeated root. Conversely, if the discriminant is negative, no real solutions exist as the roots are complex numbers (involving imaginary numbers).
In our example, the discriminant was positive (17), meaning we have two distinct real solutions.
discriminant
The discriminant plays a crucial role in understanding the nature of the solutions of a quadratic equation. It is the expression \( b^2 - 4ac \) found inside the square root of the quadratic formula. To analyze this further:
- A positive discriminant indicates that the equation has two distinct real solutions.- A discriminant equal to zero means the equation has exactly one real solution (a repeated root).- A negative discriminant means no real solutions; instead, the solutions will be complex numbers.
In the example equation \( x^2 + x - 4 \), the discriminant was calculated as 17, which is positive. This result points to the equation having two real and distinct solutions. Always remember, calculating the discriminant is the first step to predict the type of solutions you will encounter.
- A positive discriminant indicates that the equation has two distinct real solutions.- A discriminant equal to zero means the equation has exactly one real solution (a repeated root).- A negative discriminant means no real solutions; instead, the solutions will be complex numbers.
In the example equation \( x^2 + x - 4 \), the discriminant was calculated as 17, which is positive. This result points to the equation having two real and distinct solutions. Always remember, calculating the discriminant is the first step to predict the type of solutions you will encounter.
standard form of a quadratic equation
The standard form of a quadratic equation is crucial for applying the quadratic formula. It is expressed as:
This might involve rearranging terms, as seen in the example \( x^2 + x = 4 \). It was first converted to \( x^2 + x - 4 = 0 \) to fit the standard form. Once in standard form, you can easily identify \( a \), \( b \), and \( c \) to substitute into the quadratic formula. Ensuring your equation is in standard form streamlines the process of finding solutions.
- \( ax^2 + bx + c = 0 \)
This might involve rearranging terms, as seen in the example \( x^2 + x = 4 \). It was first converted to \( x^2 + x - 4 = 0 \) to fit the standard form. Once in standard form, you can easily identify \( a \), \( b \), and \( c \) to substitute into the quadratic formula. Ensuring your equation is in standard form streamlines the process of finding solutions.
Other exercises in this chapter
Problem 39
For the following exercises, perform the indicated operation and express the result as a simplified complex number. $$ i^{8} $$
View solution Problem 39
Solve for \(h : A=\frac{1}{2} h\left(b_{1}+b_{2}\right)\)
View solution Problem 39
For each of the following exercises, find and plot the \(x\) -and \(y\) -intercepts, and graph the straight line based on those two points. $$x-2 y=8$$
View solution Problem 39
\text { Solve for } h: A=\frac{1}{2} h\left(b_{1}+b_{2}\right)
View solution