Problem 15
Question
For the following problems, solve the equations. $$ a^{2}+6 a+8=0 $$
Step-by-Step Solution
Verified Answer
Question: Find the two possible values of 'a' that satisfy the equation $$a^{2}+6a+8=0$$.
Answer: The two possible values of 'a' that satisfy the given equation are $$a = -2$$ and $$a = -4$$.
1Step 1: Identify the coefficients
The given equation is $$a^{2}+6a+8=0$$. We can see that the coefficients are $$a = 1$$, $$b = 6$$, and $$c = 8$$.
2Step 2: Plug the coefficients into the quadratic formula
Now, we will plug the coefficients into the quadratic formula:
$$
a = \frac{-6 \pm \sqrt{6^2 - 4\cdot1\cdot8}}{2\cdot1}
$$
3Step 3: Simplify the equation
Let's simplify the equation:
$$
a = \frac{-6 \pm \sqrt{36 - 32}}{2}
$$
Now calculate the discriminant ($$b^2 - 4ac = 36 - 32$$):
$$
a = \frac{-6 \pm \sqrt{4}}{2}
$$
4Step 4: Solve for two possible values of 'a'
As we have a square root, there will be two possible values for $$a$$ depending on whether we use the positive or the negative square root. Let's solve for both:
For the positive square root:
$$
a_1 = \frac{-6 + 2}{2} = \frac{-4}{2} = -2
$$
For the negative square root:
$$
a_2 = \frac{-6 - 2}{2} = \frac{-8}{2} = -4
$$
The two possible values of $$a$$ that satisfy the given equation are $$a = -2$$ and $$a = -4$$.
Key Concepts
Quadratic FormulaDiscriminantSolving Equations
Quadratic Formula
The quadratic formula is a powerful method used for solving any quadratic equation of the form \(ax^2 + bx + c = 0\). This formula simplifies the process of finding the roots, or solutions, of the quadratic equation. It is expressed as:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Here's a breakdown of the formula:
- \(a\), \(b\), and \(c\) are the coefficients of the terms \(x^2\), \(x\), and the constant term, respectively.
- The \(\pm\) symbol means you'll calculate two separate solutions: one with \(+\) and one with \(-\).
- The term \(\sqrt{b^2 - 4ac}\) is crucial as it determines the nature of the roots, which we'll explore next.
Discriminant
The discriminant in a quadratic equation is found within the quadratic formula. It is the part under the square root symbol: \(b^2 - 4ac\). The discriminant indicates how many and what type of solutions exist:
- If \(b^2 - 4ac > 0\): The equation has two distinct real roots since the square root of a positive number is a real number.
- If \(b^2 - 4ac = 0\): The equation has exactly one real root, also known as a double root, because the square root of zero is zero.
- If \(b^2 - 4ac < 0\): The equation has no real roots; instead, there are two complex roots, which occur as complex conjugates.
Solving Equations
Solving quadratic equations involves finding the values of the variable that make the equation true. In our example, the equation is \(a^2 + 6a + 8 = 0\). Here, we used the quadratic formula to solve for \(a\) because the equation did not easily factorize at a glance.
Starting with identifying the coefficients as \(a = 1\), \(b = 6\), and \(c = 8\), we plugged these into the quadratic formula:\[a = \frac{-6 \pm \sqrt{36 - 32}}{2(1)}\]Simplifying, we found the roots by evaluating the square roots and performing basic arithmetic:
Starting with identifying the coefficients as \(a = 1\), \(b = 6\), and \(c = 8\), we plugged these into the quadratic formula:\[a = \frac{-6 \pm \sqrt{36 - 32}}{2(1)}\]Simplifying, we found the roots by evaluating the square roots and performing basic arithmetic:
- Using the positive square root gives \(a_1 = -2\).
- Using the negative square root gives \(a_2 = -4\).
Other exercises in this chapter
Problem 14
For the following problems, solve the equations, if possible. $$ (x-4)(x+2)=0 $$
View solution Problem 14
For the following problems, write the values of \(a, b,\) and \(c\) in quadratic equations. $$ -3 a^{2}+4 a-1=0 $$
View solution Problem 15
A woman's glasses accidently fall off her face while she is looking out of a window in a tall building. The equation relating \(h,\) the height above the ground
View solution Problem 15
Solve each quadratic equation by extraction of roots. $$ (x-5)^{2}=1 $$
View solution